diff --git a/problems/0001.两数之和.md b/problems/0001.两数之和.md index 9571a773..459a66ad 100644 --- a/problems/0001.两数之和.md +++ b/problems/0001.两数之和.md @@ -274,6 +274,24 @@ class Solution { } } ``` +C#: +```csharp +public class Solution { + public int[] TwoSum(int[] nums, int target) { + Dictionary dic= new Dictionary(); + for(int i=0;i diff --git a/problems/0019.删除链表的倒数第N个节点.md b/problems/0019.删除链表的倒数第N个节点.md index 813e9b02..c36900bc 100644 --- a/problems/0019.删除链表的倒数第N个节点.md +++ b/problems/0019.删除链表的倒数第N个节点.md @@ -39,7 +39,7 @@ 分为如下几步： -* 首先这里我推荐大家使用虚拟头结点，这样方面处理删除实际头结点的逻辑，如果虚拟头结点不清楚，可以看这篇： [链表：听说用虚拟头节点会方便很多？](https://programmercarl.com/0203.移除链表元素.html) +* 首先这里我推荐大家使用虚拟头结点，这样方便处理删除实际头结点的逻辑，如果虚拟头结点不清楚，可以看这篇： [链表：听说用虚拟头节点会方便很多？](https://programmercarl.com/0203.移除链表元素.html) * 定义fast指针和slow指针，初始值为虚拟头结点，如图： @@ -289,6 +289,28 @@ func removeNthFromEnd(_ head: ListNode?, _ n: Int) -> ListNode? { return dummyHead.next } ``` - +Scala: +```scala +object Solution { + def removeNthFromEnd(head: ListNode, n: Int): ListNode = { + val dummy = new ListNode(-1, head) // 定义虚拟头节点 + var fast = head // 快指针从头开始走 + var slow = dummy // 慢指针从虚拟头开始头 + // 因为参数 n 是不可变量，所以不能使用 while(n>0){n-=1}的方式 + for (i <- 0 until n) { + fast = fast.next + } + // 快指针和满指针一起走，直到fast走到null + while (fast != null) { + slow = slow.next + fast = fast.next + } + // 删除slow的下一个节点 + slow.next = slow.next.next + // 返回虚拟头节点的下一个 + dummy.next + } +} +``` -----------------------
diff --git a/problems/0024.两两交换链表中的节点.md b/problems/0024.两两交换链表中的节点.md index ce75e0d7..2289c229 100644 --- a/problems/0024.两两交换链表中的节点.md +++ b/problems/0024.两两交换链表中的节点.md @@ -311,7 +311,29 @@ func swapPairs(_ head: ListNode?) -> ListNode? { return dummyHead.next } ``` - +Scala: +```scala +// 虚拟头节点 +object Solution { + def swapPairs(head: ListNode): ListNode = { + var dummy = new ListNode(0, head) // 虚拟头节点 + var pre = dummy + var cur = head + // 当pre的下一个和下下个都不为空，才进行两两转换 + while (pre.next != null && pre.next.next != null) { + var tmp: ListNode = cur.next.next // 缓存下一次要进行转换的第一个节点 + pre.next = cur.next // 步骤一 + cur.next.next = cur // 步骤二 + cur.next = tmp // 步骤三 + // 下面是准备下一轮的交换 + pre = cur + cur = tmp + } + // 最终返回dummy虚拟头节点的下一个，return可以省略 + dummy.next + } +} +``` -----------------------
diff --git a/problems/0027.移除元素.md b/problems/0027.移除元素.md index 590cf0b9..4b50d666 100644 --- a/problems/0027.移除元素.md +++ b/problems/0027.移除元素.md @@ -81,7 +81,7 @@ public: **双指针法（快慢指针法）在数组和链表的操作中是非常常见的，很多考察数组、链表、字符串等操作的面试题，都使用双指针法。** -后序都会一一介绍到，本题代码如下： +后续都会一一介绍到，本题代码如下： ```CPP // 时间复杂度：O(n) @@ -328,6 +328,20 @@ int removeElement(int* nums, int numsSize, int val){ return slow; } ``` - +Scala: +```scala +object Solution { + def removeElement(nums: Array[Int], `val`: Int): Int = { + var slow = 0 + for (fast <- 0 until nums.length) { + if (`val` != nums(fast)) { + nums(slow) = nums(fast) + slow += 1 + } + } + slow + } +} +``` -----------------------
diff --git a/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md b/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md index dfd90b82..260462c2 100644 --- a/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md +++ b/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md @@ -480,7 +480,52 @@ var searchRange = function(nums, target) { return [-1, -1]; }; ``` +### Scala +```scala +object Solution { + def searchRange(nums: Array[Int], target: Int): Array[Int] = { + var left = getLeftBorder(nums, target) + var right = getRightBorder(nums, target) + if (left == -2 || right == -2) return Array(-1, -1) + if (right - left > 1) return Array(left + 1, right - 1) + Array(-1, -1) + } + // 寻找左边界 + def getLeftBorder(nums: Array[Int], target: Int): Int = { + var leftBorder = -2 + var left = 0 + var right = nums.length - 1 + while (left <= right) { + var mid = left + (right - left) / 2 + if (nums(mid) >= target) { + right = mid - 1 + leftBorder = right + } else { + left = mid + 1 + } + } + leftBorder + } + + // 寻找右边界 + def getRightBorder(nums: Array[Int], target: Int): Int = { + var rightBorder = -2 + var left = 0 + var right = nums.length - 1 + while (left <= right) { + var mid = left + (right - left) / 2 + if (nums(mid) <= target) { + left = mid + 1 + rightBorder = left + } else { + right = mid - 1 + } + } + rightBorder + } +} +``` -----------------------
diff --git a/problems/0035.搜索插入位置.md b/problems/0035.搜索插入位置.md index 8a8f9706..6c04e7de 100644 --- a/problems/0035.搜索插入位置.md +++ b/problems/0035.搜索插入位置.md @@ -316,7 +316,26 @@ func searchInsert(_ nums: [Int], _ target: Int) -> Int { return right + 1 } ``` - +### Scala +```scala +object Solution { + def searchInsert(nums: Array[Int], target: Int): Int = { + var left = 0 + var right = nums.length - 1 + while (left <= right) { + var mid = left + (right - left) / 2 + if (target == nums(mid)) { + return mid + } else if (target > nums(mid)) { + left = mid + 1 + } else { + right = mid - 1 + } + } + right + 1 + } +} +``` ### PHP diff --git a/problems/0045.跳跃游戏II.md b/problems/0045.跳跃游戏II.md index 4caff042..4e3ab24a 100644 --- a/problems/0045.跳跃游戏II.md +++ b/problems/0045.跳跃游戏II.md @@ -217,18 +217,26 @@ class Solution: ### Go ```Go func jump(nums []int) int { - dp:=make([]int ,len(nums)) - dp[0]=0 + dp := make([]int, len(nums)) + dp[0] = 0//初始第一格跳跃数一定为0 - for i:=1;ii{ - dp[i]=min(dp[j]+1,dp[i]) - } - } - } - return dp[len(nums)-1] + for i := 1; i < len(nums); i++ { + dp[i] = i + for j := 0; j < i; j++ { + if nums[j] + j >= i {//nums[j]为起点,j为往右跳的覆盖范围,这行表示从j能跳到i + dp[i] = min(dp[j] + 1, dp[i])//更新最小能到i的跳跃次数 + } + } + } + return dp[len(nums)-1] +} + +func min(a, b int) int { + if a < b { + return a + } else { + return b + } } ``` diff --git a/problems/0056.合并区间.md b/problems/0056.合并区间.md index b44d602c..92b66473 100644 --- a/problems/0056.合并区间.md +++ b/problems/0056.合并区间.md @@ -112,8 +112,8 @@ public: }; ``` -* 时间复杂度：$O(n\log n)$ ，有一个快排 -* 空间复杂度：$O(1)$，我没有算result数组（返回值所需容器占的空间） +* 时间复杂度：O(nlog n) ，有一个快排 +* 空间复杂度：O(n)，有一个快排，最差情况(倒序)时，需要n次递归调用。因此确实需要O(n)的栈空间 ## 总结 diff --git a/problems/0059.螺旋矩阵II.md b/problems/0059.螺旋矩阵II.md index 93735895..22229302 100644 --- a/problems/0059.螺旋矩阵II.md +++ b/problems/0059.螺旋矩阵II.md @@ -130,57 +130,37 @@ Java： ```Java class Solution { public int[][] generateMatrix(int n) { + int loop = 0; // 控制循环次数 int[][] res = new int[n][n]; + int start = 0; // 每次循环的开始点(start, start) + int count = 1; // 定义填充数字 + int i, j; - // 循环次数 - int loop = n / 2; - - // 定义每次循环起始位置 - int startX = 0; - int startY = 0; - - // 定义偏移量 - int offset = 1; - - // 定义填充数字 - int count = 1; - - // 定义中间位置 - int mid = n / 2; - while (loop > 0) { - int i = startX; - int j = startY; - + while (loop++ < n / 2) { // 判断边界后，loop从1开始 // 模拟上侧从左到右 - for (; j startY; j--) { + for (; j >= loop; j--) { res[i][j] = count++; } // 模拟左侧从下到上 - for (; i > startX; i--) { + for (; i >= loop; i--) { res[i][j] = count++; } - - loop--; - - startX += 1; - startY += 1; - - offset += 2; + start++; } if (n % 2 == 1) { - res[mid][mid] = count; + res[start][start] = count; } return res; @@ -564,6 +544,57 @@ int** generateMatrix(int n, int* returnSize, int** returnColumnSizes){ return ans; } ``` +Scala: +```scala +object Solution { + def generateMatrix(n: Int): Array[Array[Int]] = { + var res = Array.ofDim[Int](n, n) // 定义一个n*n的二维矩阵 + var num = 1 // 标志当前到了哪个数字 + var i = 0 // 横坐标 + var j = 0 // 竖坐标 + while (num <= n * n) { + // 向右：当j不越界，并且下一个要填的数字是空白时 + while (j < n && res(i)(j) == 0) { + res(i)(j) = num // 当前坐标等于num + num += 1 // num++ + j += 1 // 竖坐标+1 + } + i += 1 // 下移一行 + j -= 1 // 左移一列 + + // 剩下的都同上 + + // 向下 + while (i < n && res(i)(j) == 0) { + res(i)(j) = num + num += 1 + i += 1 + } + i -= 1 + j -= 1 + + // 向左 + while (j >= 0 && res(i)(j) == 0) { + res(i)(j) = num + num += 1 + j -= 1 + } + i -= 1 + j += 1 + + // 向上 + while (i >= 0 && res(i)(j) == 0) { + res(i)(j) = num + num += 1 + i -= 1 + } + i += 1 + j += 1 + } + res + } +} +``` -----------------------
diff --git a/problems/0070.爬楼梯完全背包版本.md b/problems/0070.爬楼梯完全背包版本.md index 2286de2d..0f482bb7 100644 --- a/problems/0070.爬楼梯完全背包版本.md +++ b/problems/0070.爬楼梯完全背包版本.md @@ -199,6 +199,28 @@ var climbStairs = function(n) { }; ``` +TypeScript： + +```typescript +function climbStairs(n: number): number { + const m: number = 2; // 本题m为2 + const dp: number[] = new Array(n + 1).fill(0); + dp[0] = 1; + // 遍历背包 + for (let i = 1; i <= n; i++) { + // 遍历物品 + for (let j = 1; j <= m; j++) { + if (j <= i) { + dp[i] += dp[i - j]; + } + } + } + return dp[n]; +}; +``` + + + -----------------------
diff --git a/problems/0077.组合.md b/problems/0077.组合.md index 4560c5b7..9e0398ab 100644 --- a/problems/0077.组合.md +++ b/problems/0077.组合.md @@ -27,7 +27,7 @@ 也可以直接看我的B站视频：[带你学透回溯算法-组合问题（对应力扣题目：77.组合）](https://www.bilibili.com/video/BV1ti4y1L7cv#reply3733925949) -# 思路 +## 思路 本题这是回溯法的经典题目。 @@ -232,7 +232,7 @@ void backtracking(参数) { **对比一下本题的代码，是不是发现有点像！** 所以有了这个模板，就有解题的大体方向，不至于毫无头绪。 -# 总结 +## 总结 组合问题是回溯法解决的经典问题，我们开始的时候给大家列举一个很形象的例子，就是n为100，k为50的话，直接想法就需要50层for循环。 @@ -242,7 +242,7 @@ void backtracking(参数) { 接着用回溯法三部曲，逐步分析了函数参数、终止条件和单层搜索的过程。 -# 剪枝优化 +## 剪枝优化 我们说过，回溯法虽然是暴力搜索，但也有时候可以有点剪枝优化一下的。 @@ -324,7 +324,7 @@ public: }; ``` -# 剪枝总结 +## 剪枝总结 本篇我们准对求组合问题的回溯法代码做了剪枝优化，这个优化如果不画图的话，其实不好理解，也不好讲清楚。 @@ -334,10 +334,10 @@ public: -# 其他语言版本 +## 其他语言版本 -## Java： +### Java： ```java class Solution { List> result = new ArrayList<>(); @@ -366,6 +366,8 @@ class Solution { } ``` +### Python + Python2: ```python class Solution(object): @@ -395,7 +397,6 @@ class Solution(object): return result ``` -## Python ```python class Solution: def combine(self, n: int, k: int) -> List[List[int]]: @@ -432,7 +433,7 @@ class Solution: ``` -## javascript +### javascript 剪枝： ```javascript @@ -456,7 +457,7 @@ const combineHelper = (n, k, startIndex) => { } ``` -## TypeScript +### TypeScript ```typescript function combine(n: number, k: number): number[][] { @@ -479,7 +480,7 @@ function combine(n: number, k: number): number[][] { -## Go +### Go ```Go var res [][]int func combine(n int, k int) [][]int { @@ -534,7 +535,7 @@ func backtrack(n,k,start int,track []int){ } ``` -## C +### C ```c int* path; int pathTop; @@ -642,7 +643,7 @@ int** combine(int n, int k, int* returnSize, int** returnColumnSizes){ } ``` -## Swift +### Swift ```swift func combine(_ n: Int, _ k: Int) -> [[Int]] { diff --git a/problems/0096.不同的二叉搜索树.md b/problems/0096.不同的二叉搜索树.md index 41fcb8fe..25561b50 100644 --- a/problems/0096.不同的二叉搜索树.md +++ b/problems/0096.不同的二叉搜索树.md @@ -227,7 +227,33 @@ const numTrees =(n) => { }; ``` -C: +TypeScript + +```typescript +function numTrees(n: number): number { + /** + dp[i]: i个节点对应的种树 + dp[0]: -1; 无意义； + dp[1]: 1; + ... + dp[i]: 2 * dp[i - 1] + + (dp[1] * dp[i - 2] + dp[2] * dp[i - 3] + ... + dp[i - 2] * dp[1]); 从1加到i-2 + */ + const dp: number[] = []; + dp[0] = -1; // 表示无意义 + dp[1] = 1; + for (let i = 2; i <= n; i++) { + dp[i] = 2 * dp[i - 1]; + for (let j = 1, end = i - 1; j < end; j++) { + dp[i] += dp[j] * dp[end - j]; + } + } + return dp[n]; +}; +``` + +### C + ```c //开辟dp数组 int *initDP(int n) { diff --git a/problems/0101.对称二叉树.md b/problems/0101.对称二叉树.md index e4e232c8..1eb43589 100644 --- a/problems/0101.对称二叉树.md +++ b/problems/0101.对称二叉树.md @@ -238,7 +238,7 @@ public: }; ``` -# 总结 +## 总结 这次我们又深度剖析了一道二叉树的“简单题”，大家会发现，真正的把题目搞清楚其实并不简单，leetcode上accept了和真正掌握了还是有距离的。 @@ -248,7 +248,7 @@ public: 如果已经做过这道题目的同学，读完文章可以再去看看这道题目，思考一下，会有不一样的发现！ -# 相关题目推荐 +## 相关题目推荐 这两道题目基本和本题是一样的，只要稍加修改就可以AC。 diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md index ab8f2e57..5f69f53d 100644 --- a/problems/0102.二叉树的层序遍历.md +++ b/problems/0102.二叉树的层序遍历.md @@ -82,6 +82,26 @@ public: } }; ``` +```CPP +# 递归法 +class Solution { +public: + void order(TreeNode* cur, vector>& result, int depth) + { + if (cur == nullptr) return; + if (result.size() == depth) result.push_back(vector()); + result[depth].push_back(cur->val); + order(cur->left, result, depth + 1); + order(cur->right, result, depth + 1); + } + vector> levelOrder(TreeNode* root) { + vector> result; + int depth = 0; + order(root, result, depth); + return result; + } +}; +``` python3代码： diff --git a/problems/0112.路径总和.md b/problems/0112.路径总和.md index 41463ec1..d3eec16b 100644 --- a/problems/0112.路径总和.md +++ b/problems/0112.路径总和.md @@ -377,22 +377,22 @@ class solution { ```java class solution { - public list> pathsum(treenode root, int targetsum) { - list> res = new arraylist<>(); + public List> pathsum(TreeNode root, int targetsum) { + List> res = new ArrayList<>(); if (root == null) return res; // 非空判断 - - list path = new linkedlist<>(); + + List path = new LinkedList<>(); preorderdfs(root, targetsum, res, path); return res; } - public void preorderdfs(treenode root, int targetsum, list> res, list path) { + public void preorderdfs(TreeNode root, int targetsum, List> res, List path) { path.add(root.val); // 遇到了叶子节点 if (root.left == null && root.right == null) { // 找到了和为 targetsum 的路径 if (targetsum - root.val == 0) { - res.add(new arraylist<>(path)); + res.add(new ArrayList<>(path)); } return; // 如果和不为 targetsum，返回 } @@ -1006,6 +1006,126 @@ func traversal(_ cur: TreeNode?, count: Int) { } ``` +## C +> 0112.路径总和 +递归法： +```c +bool hasPathSum(struct TreeNode* root, int targetSum){ + // 递归结束条件：若当前节点不存在，返回false + if(!root) + return false; + // 若当前节点为叶子节点，且targetSum-root的值为0。（当前路径上的节点值的和满足条件）返回true + if(!root->right && !root->left && targetSum == root->val) + return true; + + // 查看左子树和右子树的所有节点是否满足条件 + return hasPathSum(root->right, targetSum - root->val) || hasPathSum(root->left, targetSum - root->val); +} +``` + +迭代法: +```c +// 存储一个节点以及当前的和 +struct Pair { + struct TreeNode* node; + int sum; +}; + +bool hasPathSum(struct TreeNode* root, int targetSum){ + struct Pair stack[1000]; + int stackTop = 0; + + // 若root存在，则将节点和值封装成一个pair入栈 + if(root) { + struct Pair newPair = {root, root->val}; + stack[stackTop++] = newPair; + } + + // 当栈不为空时 + while(stackTop) { + // 出栈栈顶元素 + struct Pair topPair = stack[--stackTop]; + // 若栈顶元素为叶子节点，且和为targetSum时，返回true + if(!topPair.node->left && !topPair.node->right && topPair.sum == targetSum) + return true; + + // 若当前栈顶节点有左右孩子，计算和并入栈 + if(topPair.node->left) { + struct Pair newPair = {topPair.node->left, topPair.sum + topPair.node->left->val}; + stack[stackTop++] = newPair; + } + if(topPair.node->right) { + struct Pair newPair = {topPair.node->right, topPair.sum + topPair.node->right->val}; + stack[stackTop++] = newPair; + } + } + return false; +} +``` +> 0113.路径总和 II +```c +int** ret; +int* path; +int* colSize; +int retTop; +int pathTop; + +void traversal(const struct TreeNode* const node, int count) { + // 若当前节点为叶子节点 + if(!node->right && !node->left) { + // 若当前path上的节点值总和等于targetSum。 + if(count == 0) { + // 复制当前path + int *curPath = (int*)malloc(sizeof(int) * pathTop); + memcpy(curPath, path, sizeof(int) * pathTop); + // 记录当前path的长度为pathTop + colSize[retTop] = pathTop; + // 将当前path加入到ret数组中 + ret[retTop++] = curPath; + } + return; + } + + // 若节点有左/右孩子 + if(node->left) { + // 将左孩子的值加入path中 + path[pathTop++] = node->left->val; + traversal(node->left, count - node->left->val); + // 回溯 + pathTop--; + } + if(node->right) { + // 将右孩子的值加入path中 + path[pathTop++] = node->right->val; + traversal(node->right, count - node->right->val); + // 回溯 + --pathTop; + } +} + +int** pathSum(struct TreeNode* root, int targetSum, int* returnSize, int** returnColumnSizes){ + // 初始化数组 + ret = (int**)malloc(sizeof(int*) * 1000); + path = (int*)malloc(sizeof(int*) * 1000); + colSize = (int*)malloc(sizeof(int) * 1000); + retTop = pathTop = 0; + *returnSize = 0; + + // 若根节点不存在，返回空的ret + if(!root) + return ret; + // 将根节点加入到path中 + path[pathTop++] = root->val; + traversal(root, targetSum - root->val); + + // 设置返回ret数组大小，以及其中每个一维数组元素的长度 + *returnSize = retTop; + *returnColumnSizes = colSize; + + return ret; +} +``` + ----------------------- diff --git a/problems/0121.买卖股票的最佳时机.md b/problems/0121.买卖股票的最佳时机.md index f0bc3b97..a2498bb6 100644 --- a/problems/0121.买卖股票的最佳时机.md +++ b/problems/0121.买卖股票的最佳时机.md @@ -426,6 +426,46 @@ var maxProfit = function(prices) { }; ``` +TypeScript： + +> 贪心法 + +```typescript +function maxProfit(prices: number[]): number { + if (prices.length === 0) return 0; + let buy: number = prices[0]; + let profitMax: number = 0; + for (let i = 1, length = prices.length; i < length; i++) { + profitMax = Math.max(profitMax, prices[i] - buy); + buy = Math.min(prices[i], buy); + } + return profitMax; +}; +``` + +> 动态规划 + +```typescript +function maxProfit(prices: number[]): number { + /** + dp[i][0]: 第i天持有股票的最大现金 + dp[i][1]: 第i天不持有股票的最大现金 + */ + const length = prices.length; + if (length === 0) return 0; + const dp: number[][] = []; + dp[0] = [-prices[0], 0]; + for (let i = 1; i < length; i++) { + dp[i] = []; + dp[i][0] = Math.max(dp[i - 1][0], -prices[i]); + dp[i][1] = Math.max(dp[i - 1][0] + prices[i], dp[i - 1][1]); + } + return dp[length - 1][1]; +}; +``` + + + -----------------------
diff --git a/problems/0122.买卖股票的最佳时机II（动态规划）.md b/problems/0122.买卖股票的最佳时机II（动态规划）.md index 5a165a14..12b21fde 100644 --- a/problems/0122.买卖股票的最佳时机II（动态规划）.md +++ b/problems/0122.买卖股票的最佳时机II（动态规划）.md @@ -295,6 +295,42 @@ const maxProfit = (prices) => { } ``` +TypeScript： + +> 动态规划 + +```typescript +function maxProfit(prices: number[]): number { + /** + dp[i][0]: 第i天持有股票 + dp[i][1]: 第i天不持有股票 + */ + const length: number = prices.length; + if (length === 0) return 0; + const dp: number[][] = new Array(length).fill(0).map(_ => []); + dp[0] = [-prices[0], 0]; + for (let i = 1; i < length; i++) { + dp[i][0] = Math.max(dp[i - 1][0], dp[i - 1][1] - prices[i]); + dp[i][1] = Math.max(dp[i - 1][1], dp[i - 1][0] + prices[i]); + } + return dp[length - 1][1]; +}; +``` + +> 贪心法 + +```typescript +function maxProfit(prices: number[]): number { + let resProfit: number = 0; + for (let i = 1, length = prices.length; i < length; i++) { + if (prices[i] > prices[i - 1]) { + resProfit += prices[i] - prices[i - 1]; + } + } + return resProfit; +}; +``` + ----------------------- diff --git a/problems/0123.买卖股票的最佳时机III.md b/problems/0123.买卖股票的最佳时机III.md index 56ade343..67c99497 100644 --- a/problems/0123.买卖股票的最佳时机III.md +++ b/problems/0123.买卖股票的最佳时机III.md @@ -352,6 +352,36 @@ const maxProfit = prices => { }; ``` +TypeScript： + +> 版本一 + +```typescript +function maxProfit(prices: number[]): number { + /** + dp[i][0]: 无操作; + dp[i][1]: 第一次买入； + dp[i][2]: 第一次卖出； + dp[i][3]: 第二次买入； + dp[i][4]: 第二次卖出； + */ + const length: number = prices.length; + if (length === 0) return 0; + const dp: number[][] = new Array(length).fill(0) + .map(_ => new Array(5).fill(0)); + dp[0][1] = -prices[0]; + dp[0][3] = -prices[0]; + for (let i = 1; i < length; i++) { + dp[i][0] = dp[i - 1][0]; + dp[i][1] = Math.max(dp[i - 1][1], -prices[i]); + dp[i][2] = Math.max(dp[i - 1][2], dp[i - 1][1] + prices[i]); + dp[i][3] = Math.max(dp[i - 1][3], dp[i - 1][2] - prices[i]); + dp[i][4] = Math.max(dp[i - 1][4], dp[i - 1][3] + prices[i]); + } + return Math.max(dp[length - 1][2], dp[length - 1][4]); +}; +``` + Go: > 版本一： diff --git a/problems/0139.单词拆分.md b/problems/0139.单词拆分.md index ac834f04..5b4e92b9 100644 --- a/problems/0139.单词拆分.md +++ b/problems/0139.单词拆分.md @@ -345,6 +345,48 @@ const wordBreak = (s, wordDict) => { } ``` +TypeScript： + +> 动态规划 + +```typescript +function wordBreak(s: string, wordDict: string[]): boolean { + const dp: boolean[] = new Array(s.length + 1).fill(false); + dp[0] = true; + for (let i = 1; i <= s.length; i++) { + for (let j = 0; j < i; j++) { + const tempStr: string = s.slice(j, i); + if (wordDict.includes(tempStr) && dp[j] === true) { + dp[i] = true; + break; + } + } + } + return dp[s.length]; +}; +``` + +> 记忆化回溯 + +```typescript +function wordBreak(s: string, wordDict: string[]): boolean { + // 只需要记忆结果为false的情况 + const memory: boolean[] = []; + return backTracking(s, wordDict, 0, memory); + function backTracking(s: string, wordDict: string[], startIndex: number, memory: boolean[]): boolean { + if (startIndex >= s.length) return true; + if (memory[startIndex] === false) return false; + for (let i = startIndex + 1, length = s.length; i <= length; i++) { + const str: string = s.slice(startIndex, i); + if (wordDict.includes(str) && backTracking(s, wordDict, i, memory)) + return true; + } + memory[startIndex] = false; + return false; + } +}; +``` + ----------------------- diff --git a/problems/0142.环形链表II.md b/problems/0142.环形链表II.md index e8ca950d..f8e62d45 100644 --- a/problems/0142.环形链表II.md +++ b/problems/0142.环形链表II.md @@ -370,7 +370,31 @@ ListNode *detectCycle(ListNode *head) { } ``` - +Scala: +```scala +object Solution { + def detectCycle(head: ListNode): ListNode = { + var fast = head // 快指针 + var slow = head // 慢指针 + while (fast != null && fast.next != null) { + fast = fast.next.next // 快指针一次走两步 + slow = slow.next // 慢指针一次走一步 + // 如果相遇，fast快指针回到头 + if (fast == slow) { + fast = head + // 两个指针一步一步的走，第一次相遇的节点必是入环节点 + while (fast != slow) { + fast = fast.next + slow = slow.next + } + return fast + } + } + // 如果fast指向空值，必然无环返回null + null + } +} +``` -----------------------
diff --git a/problems/0150.逆波兰表达式求值.md b/problems/0150.逆波兰表达式求值.md index fd3d69aa..6ce7e2f9 100644 --- a/problems/0150.逆波兰表达式求值.md +++ b/problems/0150.逆波兰表达式求值.md @@ -136,19 +136,19 @@ java: class Solution { public int evalRPN(String[] tokens) { Deque stack = new LinkedList(); - for (int i = 0; i < tokens.length; ++i) { - if ("+".equals(tokens[i])) { // leetcode 内置jdk的问题，不能使用==判断字符串是否相等 + for (String s : tokens) { + if ("+".equals(s)) { // leetcode 内置jdk的问题，不能使用==判断字符串是否相等 stack.push(stack.pop() + stack.pop()); // 注意 - 和/ 需要特殊处理 - } else if ("-".equals(tokens[i])) { + } else if ("-".equals(s)) { stack.push(-stack.pop() + stack.pop()); - } else if ("*".equals(tokens[i])) { + } else if ("*".equals(s)) { stack.push(stack.pop() * stack.pop()); - } else if ("/".equals(tokens[i])) { + } else if ("/".equals(s)) { int temp1 = stack.pop(); int temp2 = stack.pop(); stack.push(temp2 / temp1); } else { - stack.push(Integer.valueOf(tokens[i])); + stack.push(Integer.valueOf(s)); } } return stack.pop(); diff --git a/problems/0188.买卖股票的最佳时机IV.md b/problems/0188.买卖股票的最佳时机IV.md index 61c558a1..27eb38c3 100644 --- a/problems/0188.买卖股票的最佳时机IV.md +++ b/problems/0188.买卖股票的最佳时机IV.md @@ -409,5 +409,27 @@ var maxProfit = function(k, prices) { }; ``` +TypeScript： + +```typescript +function maxProfit(k: number, prices: number[]): number { + const length: number = prices.length; + if (length === 0) return 0; + const dp: number[][] = new Array(length).fill(0) + .map(_ => new Array(k * 2 + 1).fill(0)); + for (let i = 1; i <= k; i++) { + dp[0][i * 2 - 1] = -prices[0]; + } + for (let i = 1; i < length; i++) { + for (let j = 1; j < 2 * k + 1; j++) { + dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - 1] + Math.pow(-1, j) * prices[i]); + } + } + return dp[length - 1][2 * k]; +}; +``` + + + -----------------------
diff --git a/problems/0198.打家劫舍.md b/problems/0198.打家劫舍.md index dfe1f3a0..a828b9a9 100644 --- a/problems/0198.打家劫舍.md +++ b/problems/0198.打家劫舍.md @@ -189,6 +189,29 @@ const rob = nums => { }; ``` +TypeScript： + +```typescript +function rob(nums: number[]): number { + /** + dp[i]: 前i个房屋能偷到的最大金额 + dp[0]: nums[0]; + dp[1]: max(nums[0], nums[1]); + ... + dp[i]: max(dp[i-1], dp[i-2]+nums[i]); + */ + const length: number = nums.length; + if (length === 1) return nums[0]; + const dp: number[] = []; + dp[0] = nums[0]; + dp[1] = Math.max(nums[0], nums[1]); + for (let i = 2; i < length; i++) { + dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i]); + } + return dp[length - 1]; +}; +``` + diff --git a/problems/0202.快乐数.md b/problems/0202.快乐数.md index 741a735a..be8686f7 100644 --- a/problems/0202.快乐数.md +++ b/problems/0202.快乐数.md @@ -385,5 +385,61 @@ bool isHappy(int n){ return bHappy; } ``` + +Scala: +```scala +object Solution { + // 引入mutable + import scala.collection.mutable + def isHappy(n: Int): Boolean = { + // 存放每次计算后的结果 + val set: mutable.HashSet[Int] = new mutable.HashSet[Int]() + var tmp = n // 因为形参是不可变量，所以需要找到一个临时变量 + // 开始进入循环 + while (true) { + val sum = getSum(tmp) // 获取这个数每个值的平方和 + if (sum == 1) return true // 如果最终等于 1，则返回true + // 如果set里面已经有这个值了，说明进入无限循环，可以返回false，否则添加这个值到set + if (set.contains(sum)) return false + else set.add(sum) + tmp = sum + } + // 最终需要返回值，直接返回个false + false + } + + def getSum(n: Int): Int = { + var sum = 0 + var tmp = n + while (tmp != 0) { + sum += (tmp % 10) * (tmp % 10) + tmp = tmp / 10 + } + sum + } + + +C#： +```csharp +public class Solution { + private int getSum(int n) { + int sum = 0; + //每位数的换算 + while (n > 0) { + sum += (n % 10) * (n % 10); + n /= 10; + } + return sum; + } + public bool IsHappy(int n) { + HashSet set = new HashSet(); + while(n != 1 && !set.Contains(n)) { //判断避免循环 + set.Add(n); + n = getSum(n); + } + return n == 1; + } +} +``` -----------------------
diff --git a/problems/0203.移除链表元素.md b/problems/0203.移除链表元素.md index 751553e2..67776529 100644 --- a/problems/0203.移除链表元素.md +++ b/problems/0203.移除链表元素.md @@ -478,6 +478,36 @@ impl Solution { } } ``` - +Scala: +```scala +/** + * Definition for singly-linked list. + * class ListNode(_x: Int = 0, _next: ListNode = null) { + * var next: ListNode = _next + * var x: Int = _x + * } + */ +object Solution { + def removeElements(head: ListNode, `val`: Int): ListNode = { + if (head == null) return head + var dummy = new ListNode(-1, head) // 定义虚拟头节点 + var cur = head // cur 表示当前节点 + var pre = dummy // pre 表示cur前一个节点 + while (cur != null) { + if (cur.x == `val`) { + // 相等，就删除那么cur的前一个节点pre执行cur的下一个 + pre.next = cur.next + } else { + // 不相等，pre就等于当前cur节点 + pre = cur + } + // 向下迭代 + cur = cur.next + } + // 最终返回dummy的下一个，就是链表的头 + dummy.next + } +} +``` -----------------------
diff --git a/problems/0206.翻转链表.md b/problems/0206.翻转链表.md index 941928ba..25b16907 100644 --- a/problems/0206.翻转链表.md +++ b/problems/0206.翻转链表.md @@ -496,6 +496,40 @@ struct ListNode* reverseList(struct ListNode* head){ return reverse(NULL, head); } ``` +Scala: +双指针法： +```scala +object Solution { + def reverseList(head: ListNode): ListNode = { + var pre: ListNode = null + var cur = head + while (cur != null) { + var tmp = cur.next + cur.next = pre + pre = cur + cur = tmp + } + pre + } +} +``` +递归法： +```scala +object Solution { + def reverseList(head: ListNode): ListNode = { + reverse(null, head) + } + + def reverse(pre: ListNode, cur: ListNode): ListNode = { + if (cur == null) { + return pre // 如果当前cur为空，则返回pre + } + val tmp: ListNode = cur.next + cur.next = pre + reverse(cur, tmp) // 此时cur成为前一个节点，tmp是当前节点 + } +} +``` -----------------------
diff --git a/problems/0209.长度最小的子数组.md b/problems/0209.长度最小的子数组.md index d31cba3f..fbef7692 100644 --- a/problems/0209.长度最小的子数组.md +++ b/problems/0209.长度最小的子数组.md @@ -400,6 +400,54 @@ class Solution { } } ``` +Scala: + +滑动窗口: +```scala +object Solution { + def minSubArrayLen(target: Int, nums: Array[Int]): Int = { + var result = Int.MaxValue // 返回结果，默认最大值 + var left = 0 // 慢指针，当sum>=target，向右移动 + var sum = 0 // 窗口值的总和 + for (right <- 0 until nums.length) { + sum += nums(right) + while (sum >= target) { + result = math.min(result, right - left + 1) // 产生新结果 + sum -= nums(left) // 左指针移动，窗口总和减去左指针的值 + left += 1 // 左指针向右移动 + } + } + // 相当于三元运算符，return关键字可以省略 + if (result == Int.MaxValue) 0 else result + } +} +``` + +暴力解法: +```scala +object Solution { + def minSubArrayLen(target: Int, nums: Array[Int]): Int = { + import scala.util.control.Breaks + var res = Int.MaxValue + var subLength = 0 + for (i <- 0 until nums.length) { + var sum = 0 + Breaks.breakable( + for (j <- i until nums.length) { + sum += nums(j) + if (sum >= target) { + subLength = j - i + 1 + res = math.min(subLength, res) + Breaks.break() + } + } + ) + } + // 相当于三元运算符 + if (res == Int.MaxValue) 0 else res + } +} +``` -----------------------
diff --git a/problems/0213.打家劫舍II.md b/problems/0213.打家劫舍II.md index 8e569e46..9e698d01 100644 --- a/problems/0213.打家劫舍II.md +++ b/problems/0213.打家劫舍II.md @@ -165,7 +165,30 @@ const robRange = (nums, start, end) => { return dp[end] } ``` +TypeScript： + +```typescript +function rob(nums: number[]): number { + const length: number = nums.length; + if (length === 0) return 0; + if (length === 1) return nums[0]; + return Math.max(robRange(nums, 0, length - 2), + robRange(nums, 1, length - 1)); +}; +function robRange(nums: number[], start: number, end: number): number { + if (start === end) return nums[start]; + const dp: number[] = []; + dp[start] = nums[start]; + dp[start + 1] = Math.max(nums[start], nums[start + 1]); + for (let i = start + 2; i <= end; i++) { + dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i]); + } + return dp[end]; +} +``` + Go： + ```go // 打家劫舍Ⅱ 动态规划 // 时间复杂度O(n) 空间复杂度O(n) diff --git a/problems/0242.有效的字母异位词.md b/problems/0242.有效的字母异位词.md index 080166fd..8fd9c604 100644 --- a/problems/0242.有效的字母异位词.md +++ b/problems/0242.有效的字母异位词.md @@ -307,6 +307,52 @@ impl Solution { } } ``` + + +Scala: +```scala +object Solution { + def isAnagram(s: String, t: String): Boolean = { + // 如果两个字符串的长度不等，直接返回false + if (s.length != t.length) return false + val record = new Array[Int](26) // 记录每个单词出现了多少次 + // 遍历字符串，对于s字符串单词对应的记录+=1，t字符串对应的记录-=1 + for (i <- 0 until s.length) { + record(s(i) - 97) += 1 + record(t(i) - 97) -= 1 + } + // 如果不等于则直接返回false + for (i <- 0 until 26) { + if (record(i) != 0) { + return false + } + } + // 如果前面不返回false，说明匹配成功，返回true，return可以省略 + true + } +} +``` + + +C#： +```csharp + public bool IsAnagram(string s, string t) { + int sl=s.Length,tl=t.Length; + if(sl!=tl) return false; + int[] a = new int[26]; + for(int i = 0; i < sl; i++){ + a[s[i] - 'a']++; + a[t[i] - 'a']--; + } + foreach (int i in a) + { + if (i != 0) + return false; + } + return true; + } +``` + ## 相关题目 * 383.赎金信 diff --git a/problems/0279.完全平方数.md b/problems/0279.完全平方数.md index 9bad2085..5b15639c 100644 --- a/problems/0279.完全平方数.md +++ b/problems/0279.完全平方数.md @@ -355,5 +355,24 @@ var numSquares2 = function(n) { }; ``` +TypeScript： + +```typescript +function numSquares(n: number): number { + const goodsNum: number = Math.floor(Math.sqrt(n)); + const dp: number[] = new Array(n + 1).fill(Infinity); + dp[0] = 0; + for (let i = 1; i <= goodsNum; i++) { + const tempVal: number = i * i; + for (let j = tempVal; j <= n; j++) { + dp[j] = Math.min(dp[j], dp[j - tempVal] + 1); + } + } + return dp[n]; +}; +``` + + + -----------------------
diff --git a/problems/0309.最佳买卖股票时机含冷冻期.md b/problems/0309.最佳买卖股票时机含冷冻期.md index f3e7541b..f037fe85 100644 --- a/problems/0309.最佳买卖股票时机含冷冻期.md +++ b/problems/0309.最佳买卖股票时机含冷冻期.md @@ -325,6 +325,66 @@ const maxProfit = (prices) => { }; ``` +TypeScript： + +> 版本一，与本文思路一致 + +```typescript +function maxProfit(prices: number[]): number { + /** + dp[i][0]: 持股状态； + dp[i][1]: 无股状态，当天为非冷冻期； + dp[i][2]: 无股状态，当天卖出； + dp[i][3]: 无股状态，当天为冷冻期； + */ + const length: number = prices.length; + const dp: number[][] = new Array(length).fill(0).map(_ => []); + dp[0][0] = -prices[0]; + dp[0][1] = dp[0][2] = dp[0][3] = 0; + for (let i = 1; i < length; i++) { + dp[i][0] = Math.max( + dp[i - 1][0], + Math.max(dp[i - 1][1], dp[i - 1][3]) - prices[i] + ); + dp[i][1] = Math.max(dp[i - 1][1], dp[i - 1][3]); + dp[i][2] = dp[i - 1][0] + prices[i]; + dp[i][3] = dp[i - 1][2]; + } + const lastEl: number[] = dp[length - 1]; + return Math.max(lastEl[1], lastEl[2], lastEl[3]); +}; +``` + +> 版本二，状态定义略有不同，可以帮助理解 + +```typescript +function maxProfit(prices: number[]): number { + /** + dp[i][0]: 持股状态，当天买入； + dp[i][1]: 持股状态，当天未买入； + dp[i][2]: 无股状态，当天卖出； + dp[i][3]: 无股状态，当天未卖出； + + 买入有冷冻期限制，其实就是状态[0]只能由前一天的状态[3]得到； + 如果卖出有冷冻期限制，其实就是[2]由[1]得到。 + */ + const length: number = prices.length; + const dp: number[][] = new Array(length).fill(0).map(_ => []); + dp[0][0] = -prices[0]; + dp[0][1] = -Infinity; + dp[0][2] = dp[0][3] = 0; + for (let i = 1; i < length; i++) { + dp[i][0] = dp[i - 1][3] - prices[i]; + dp[i][1] = Math.max(dp[i - 1][1], dp[i - 1][0]); + dp[i][2] = Math.max(dp[i - 1][0], dp[i - 1][1]) + prices[i]; + dp[i][3] = Math.max(dp[i - 1][3], dp[i - 1][2]); + } + return Math.max(dp[length - 1][2], dp[length - 1][3]); +}; +``` + + + -----------------------
diff --git a/problems/0322.零钱兑换.md b/problems/0322.零钱兑换.md index 43c735be..fc0490c8 100644 --- a/problems/0322.零钱兑换.md +++ b/problems/0322.零钱兑换.md @@ -340,7 +340,21 @@ const coinChange = (coins, amount) => { } ``` +TypeScript： +```typescript +function coinChange(coins: number[], amount: number): number { + const dp: number[] = new Array(amount + 1).fill(Infinity); + dp[0] = 0; + for (let i = 0; i < coins.length; i++) { + for (let j = coins[i]; j <= amount; j++) { + if (dp[j - coins[i]] === Infinity) continue; + dp[j] = Math.min(dp[j], dp[j - coins[i]] + 1); + } + } + return dp[amount] === Infinity ? -1 : dp[amount]; +}; +``` -----------------------
diff --git a/problems/0337.打家劫舍III.md b/problems/0337.打家劫舍III.md index a4d8f6b2..6f50723d 100644 --- a/problems/0337.打家劫舍III.md +++ b/problems/0337.打家劫舍III.md @@ -429,7 +429,50 @@ const rob = root => { }; ``` +### TypeScript + +> 记忆化后序遍历 + +```typescript +const memory: Map = new Map(); +function rob(root: TreeNode | null): number { + if (root === null) return 0; + if (memory.has(root)) return memory.get(root); + // 不取当前节点 + const res1: number = rob(root.left) + rob(root.right); + // 取当前节点 + let res2: number = root.val; + if (root.left !== null) res2 += rob(root.left.left) + rob(root.left.right); + if (root.right !== null) res2 += rob(root.right.left) + rob(root.right.right); + const res: number = Math.max(res1, res2); + memory.set(root, res); + return res; +}; +``` + +> 状态标记化后序遍历 + +```typescript +function rob(root: TreeNode | null): number { + return Math.max(...robNode(root)); +}; +// [0]-不偷当前节点能获得的最大金额; [1]-偷~~ +type MaxValueArr = [number, number]; +function robNode(node: TreeNode | null): MaxValueArr { + if (node === null) return [0, 0]; + const leftArr: MaxValueArr = robNode(node.left); + const rightArr: MaxValueArr = robNode(node.right); + // 不偷 + const val1: number = Math.max(leftArr[0], leftArr[1]) + + Math.max(rightArr[0], rightArr[1]); + // 偷 + const val2: number = leftArr[0] + rightArr[0] + node.val; + return [val1, val2]; +} +``` + ### Go + ```go // 打家劫舍Ⅲ 动态规划 // 时间复杂度O(n) 空间复杂度O(logn) diff --git a/problems/0343.整数拆分.md b/problems/0343.整数拆分.md index 4a7ba6ab..279f1d71 100644 --- a/problems/0343.整数拆分.md +++ b/problems/0343.整数拆分.md @@ -274,7 +274,33 @@ var integerBreak = function(n) { }; ``` -C: +### TypeScript + +```typescript +function integerBreak(n: number): number { + /** + dp[i]: i对应的最大乘积 + dp[2]: 1; + ... + dp[i]: max( + 1 * dp[i - 1], 1 * (i - 1), + 2 * dp[i - 2], 2 * (i - 2), + ..., (i - 2) * dp[2], (i - 2) * 2 + ); + */ + const dp: number[] = new Array(n + 1).fill(0); + dp[2] = 1; + for (let i = 3; i <= n; i++) { + for (let j = 1; j <= i - 2; j++) { + dp[i] = Math.max(dp[i], j * dp[i - j], j * (i - j)); + } + } + return dp[n]; +}; +``` + +### C + ```c //初始化DP数组 int *initDP(int num) { diff --git a/problems/0349.两个数组的交集.md b/problems/0349.两个数组的交集.md index 45f19b6e..f7dab3d7 100644 --- a/problems/0349.两个数组的交集.md +++ b/problems/0349.两个数组的交集.md @@ -313,6 +313,69 @@ int* intersection1(int* nums1, int nums1Size, int* nums2, int nums2Size, int* re } ``` +Scala: + +正常解法: +```scala +object Solution { + def intersection(nums1: Array[Int], nums2: Array[Int]): Array[Int] = { + // 导入mutable + import scala.collection.mutable + // 临时Set，用于记录数组1出现的每个元素 + val tmpSet: mutable.HashSet[Int] = new mutable.HashSet[Int]() + // 结果Set，存储最终结果 + val resSet: mutable.HashSet[Int] = new mutable.HashSet[Int]() + // 遍历nums1，把每个元素添加到tmpSet + nums1.foreach(tmpSet.add(_)) + // 遍历nums2，如果在tmpSet存在就添加到resSet + nums2.foreach(elem => { + if (tmpSet.contains(elem)) { + resSet.add(elem) + } + }) + // 将结果转换为Array返回，return可以省略 + resSet.toArray + } +} +``` +骚操作1: +```scala +object Solution { + def intersection(nums1: Array[Int], nums2: Array[Int]): Array[Int] = { + // 先转为Set，然后取交集，最后转换为Array + (nums1.toSet).intersect(nums2.toSet).toArray + } +} +``` +骚操作2: +```scala +object Solution { + def intersection(nums1: Array[Int], nums2: Array[Int]): Array[Int] = { + // distinct去重，然后取交集 + (nums1.distinct).intersect(nums2.distinct) + } +} + + +C#: +```csharp + public int[] Intersection(int[] nums1, int[] nums2) { + if(nums1==null||nums1.Length==0||nums2==null||nums1.Length==0) + return new int[0]; //注意数组条件 + HashSet one = Insert(nums1); + HashSet two = Insert(nums2); + one.IntersectWith(two); + return one.ToArray(); + } + public HashSet Insert(int[] nums){ + HashSet one = new HashSet(); + foreach(int num in nums){ + one.Add(num); + } + return one; + } + +``` ## 相关题目 * 350.两个数组的交集 II diff --git a/problems/0377.组合总和Ⅳ.md b/problems/0377.组合总和Ⅳ.md index aaf27e61..1d808a3a 100644 --- a/problems/0377.组合总和Ⅳ.md +++ b/problems/0377.组合总和Ⅳ.md @@ -221,7 +221,27 @@ const combinationSum4 = (nums, target) => { }; ``` +TypeScript： + +```typescript +function combinationSum4(nums: number[], target: number): number { + const dp: number[] = new Array(target + 1).fill(0); + dp[0] = 1; + // 遍历背包 + for (let i = 1; i <= target; i++) { + // 遍历物品 + for (let j = 0, length = nums.length; j < length; j++) { + if (i >= nums[j]) { + dp[i] += dp[i - nums[j]]; + } + } + } + return dp[target]; +}; +``` + Rust + ```Rust impl Solution { pub fn combination_sum4(nums: Vec, target: i32) -> i32 { diff --git a/problems/0383.赎金信.md b/problems/0383.赎金信.md index 6177cc41..933a7c31 100644 --- a/problems/0383.赎金信.md +++ b/problems/0383.赎金信.md @@ -363,5 +363,22 @@ impl Solution { } ``` +C#： +```csharp +public bool CanConstruct(string ransomNote, string magazine) { + if(ransomNote.Length > magazine.Length) return false; + int[] letters = new int[26]; + foreach(char c in magazine){ + letters[c-'a']++; + } + foreach(char c in ransomNote){ + letters[c-'a']--; + if(letters[c-'a']<0){ + return false; + } + } + return true; + } +``` -----------------------
diff --git a/problems/0416.分割等和子集.md b/problems/0416.分割等和子集.md index 6e93ae8e..eb6601e1 100644 --- a/problems/0416.分割等和子集.md +++ b/problems/0416.分割等和子集.md @@ -417,6 +417,163 @@ var canPartition = function(nums) { ``` +C: +二维dp： +```c +/** +1. dp数组含义：dp[i][j]为背包重量为j时，从[0-i]元素和最大值 +2. 递推公式：dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - nums[i]] + nums[i]) +3. 初始化：dp[i][0]初始化为0。因为背包重量为0时，不可能放入元素。dp[0][j] = nums[0]，当j >= nums[0] && j < target时 +4. 遍历顺序：先遍历物品，再遍历背包 +*/ +#define MAX(a, b) (((a) > (b)) ? (a) : (b)) + +int getSum(int* nums, int numsSize) { + int sum = 0; + + int i; + for(i = 0; i < numsSize; ++i) { + sum += nums[i]; + } + return sum; +} + +bool canPartition(int* nums, int numsSize){ + // 求出元素总和 + int sum = getSum(nums, numsSize); + // 若元素总和为奇数，则不可能得到两个和相等的子数组 + if(sum % 2) + return false; + + // 若子数组的和等于target，则nums可以被分割 + int target = sum / 2; + // 初始化dp数组 + int dp[numsSize][target + 1]; + // dp[j][0]都应被设置为0。因为当背包重量为0时，不可放入元素 + memset(dp, 0, sizeof(int) * numsSize * (target + 1)); + + int i, j; + // 当背包重量j大于nums[0]时，可以在dp[0][j]中放入元素nums[0] + for(j = nums[0]; j <= target; ++j) { + dp[0][j] = nums[0]; + } + + for(i = 1; i < numsSize; ++i) { + for(j = 1; j <= target; ++j) { + // 若当前背包重量j小于nums[i]，则其值等于只考虑0到i-1物品时的值 + if(j < nums[i]) + dp[i][j] = dp[i - 1][j]; + // 否则，背包重量等于在背包中放入num[i]/不放入nums[i]的较大值 + else + dp[i][j] = MAX(dp[i - 1][j], dp[i - 1][j - nums[i]] + nums[i]); + } + } + // 判断背包重量为target，且考虑到所有物品时，放入的元素和是否等于target + return dp[numsSize - 1][target] == target; +} +``` +滚动数组： +```c +/** +1. dp数组含义：dp[j]为背包重量为j时，其中可放入元素的最大值 +2. 递推公式：dp[j] = max(dp[j], dp[j - nums[i]] + nums[i]) +3. 初始化：均初始化为0即可 +4. 遍历顺序：先遍历物品，再后序遍历背包 +*/ +#define MAX(a, b) (((a) > (b)) ? (a) : (b)) + +int getSum(int* nums, int numsSize) { + int sum = 0; + + int i; + for(i = 0; i < numsSize; ++i) { + sum += nums[i]; + } + return sum; +} + +bool canPartition(int* nums, int numsSize){ + // 求出元素总和 + int sum = getSum(nums, numsSize); + // 若元素总和为奇数，则不可能得到两个和相等的子数组 + if(sum % 2) + return false; + // 背包容量 + int target = sum / 2; + + // 初始化dp数组，元素均为0 + int dp[target + 1]; + memset(dp, 0, sizeof(int) * (target + 1)); + + int i, j; + // 先遍历物品，后遍历背包 + for(i = 0; i < numsSize; ++i) { + for(j = target; j >= nums[i]; --j) { + dp[j] = MAX(dp[j], dp[j - nums[i]] + nums[i]); + } + } + + // 查看背包容量为target时，元素总和是否等于target + return dp[target] == target; +} +``` + +TypeScript: + +> 一维数组，简洁 + +```typescript +function canPartition(nums: number[]): boolean { + const sum: number = nums.reduce((pre, cur) => pre + cur); + if (sum % 2 === 1) return false; + const bagSize: number = sum / 2; + const goodsNum: number = nums.length; + const dp: number[] = new Array(bagSize + 1).fill(0); + for (let i = 0; i < goodsNum; i++) { + for (let j = bagSize; j >= nums[i]; j--) { + dp[j] = Math.max(dp[j], dp[j - nums[i]] + nums[i]); + } + } + return dp[bagSize] === bagSize; +}; +``` + +> 二维数组，易懂 + +```typescript +function canPartition(nums: number[]): boolean { + /** + weightArr = nums; + valueArr = nums; + bagSize = sum / 2; (sum为nums各元素总和); + 按照0-1背包处理 + */ + const sum: number = nums.reduce((pre, cur) => pre + cur); + if (sum % 2 === 1) return false; + const bagSize: number = sum / 2; + const weightArr: number[] = nums; + const valueArr: number[] = nums; + const goodsNum: number = weightArr.length; + const dp: number[][] = new Array(goodsNum) + .fill(0) + .map(_ => new Array(bagSize + 1).fill(0)); + for (let i = weightArr[0]; i <= bagSize; i++) { + dp[0][i] = valueArr[0]; + } + for (let i = 1; i < goodsNum; i++) { + for (let j = 0; j <= bagSize; j++) { + if (j < weightArr[i]) { + dp[i][j] = dp[i - 1][j]; + } else { + dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - weightArr[i]] + valueArr[i]); + } + } + } + return dp[goodsNum - 1][bagSize] === bagSize; +}; +``` + + ----------------------- diff --git a/problems/0435.无重叠区间.md b/problems/0435.无重叠区间.md index dc89d80b..66aa1244 100644 --- a/problems/0435.无重叠区间.md +++ b/problems/0435.无重叠区间.md @@ -93,7 +93,7 @@ public: }; ``` * 时间复杂度：O(nlog n) ，有一个快排 -* 空间复杂度：O(1) +* 空间复杂度：O(n)，有一个快排，最差情况(倒序)时，需要n次递归调用。因此确实需要O(n)的栈空间 大家此时会发现如此复杂的一个问题，代码实现却这么简单！ diff --git a/problems/0452.用最少数量的箭引爆气球.md b/problems/0452.用最少数量的箭引爆气球.md index 2ab14b61..d4bbe961 100644 --- a/problems/0452.用最少数量的箭引爆气球.md +++ b/problems/0452.用最少数量的箭引爆气球.md @@ -105,8 +105,8 @@ public: }; ``` -* 时间复杂度：$O(n\log n)$，因为有一个快排 -* 空间复杂度：$O(1)$ +* 时间复杂度：O(nlog n)，因为有一个快排 +* 空间复杂度：O(1)，有一个快排，最差情况(倒序)时，需要n次递归调用。因此确实需要O(n)的栈空间 可以看出代码并不复杂。 diff --git a/problems/0454.四数相加II.md b/problems/0454.四数相加II.md index a6cd413b..962fe7a5 100644 --- a/problems/0454.四数相加II.md +++ b/problems/0454.四数相加II.md @@ -318,5 +318,32 @@ impl Solution { } ``` +C#： +```csharp +public int FourSumCount(int[] nums1, int[] nums2, int[] nums3, int[] nums4) { + Dictionary dic = new Dictionary(); + foreach(var i in nums1){ + foreach(var j in nums2){ + int sum = i + j; + if(dic.ContainsKey(sum)){ + dic[sum]++; + }else{ + dic.Add(sum, 1); + } + + } + } + int res = 0; + foreach(var a in nums3){ + foreach(var b in nums4){ + int sum = a+b; + if(dic.TryGetValue(-sum, out var result)){ + res += result; + } + } + } + return res; + } +``` -----------------------
diff --git a/problems/0474.一和零.md b/problems/0474.一和零.md index 964df4a8..d38ce03f 100644 --- a/problems/0474.一和零.md +++ b/problems/0474.一和零.md @@ -323,6 +323,129 @@ const findMaxForm = (strs, m, n) => { }; ``` +TypeScript： + +> 滚动数组，二维数组法 + +```typescript +type BinaryInfo = { numOfZero: number, numOfOne: number }; +function findMaxForm(strs: string[], m: number, n: number): number { + const goodsNum: number = strs.length; + const dp: number[][] = new Array(m + 1).fill(0) + .map(_ => new Array(n + 1).fill(0)); + for (let i = 0; i < goodsNum; i++) { + const { numOfZero, numOfOne } = countBinary(strs[i]); + for (let j = m; j >= numOfZero; j--) { + for (let k = n; k >= numOfOne; k--) { + dp[j][k] = Math.max(dp[j][k], dp[j - numOfZero][k - numOfOne] + 1); + } + } + } + return dp[m][n]; +}; +function countBinary(str: string): BinaryInfo { + let numOfZero: number = 0, + numOfOne: number = 0; + for (let s of str) { + if (s === '0') { + numOfZero++; + } else { + numOfOne++; + } + } + return { numOfZero, numOfOne }; +} +``` + +> 传统背包，三维数组法 + +```typescript +type BinaryInfo = { numOfZero: number, numOfOne: number }; +function findMaxForm(strs: string[], m: number, n: number): number { + /** + dp[i][j][k]: 前i个物品中, 背包的0容量为j, 1容量为k, 最多能放的物品数量 + */ + const goodsNum: number = strs.length; + const dp: number[][][] = new Array(goodsNum).fill(0) + .map(_ => new Array(m + 1) + .fill(0) + .map(_ => new Array(n + 1).fill(0)) + ); + const { numOfZero, numOfOne } = countBinary(strs[0]); + for (let i = numOfZero; i <= m; i++) { + for (let j = numOfOne; j <= n; j++) { + dp[0][i][j] = 1; + } + } + for (let i = 1; i < goodsNum; i++) { + const { numOfZero, numOfOne } = countBinary(strs[i]); + for (let j = 0; j <= m; j++) { + for (let k = 0; k <= n; k++) { + if (j < numOfZero || k < numOfOne) { + dp[i][j][k] = dp[i - 1][j][k]; + } else { + dp[i][j][k] = Math.max(dp[i - 1][j][k], dp[i - 1][j - numOfZero][k - numOfOne] + 1); + } + } + } + } + return dp[dp.length - 1][m][n]; +}; +function countBinary(str: string): BinaryInfo { + let numOfZero: number = 0, + numOfOne: number = 0; + for (let s of str) { + if (s === '0') { + numOfZero++; + } else { + numOfOne++; + } + } + return { numOfZero, numOfOne }; +} +``` + +> 回溯法（会超时） + +```typescript +function findMaxForm(strs: string[], m: number, n: number): number { + /** + 思路：暴力枚举strs的所有子集，记录符合条件子集的最大长度 + */ + let resMax: number = 0; + backTrack(strs, m, n, 0, []); + return resMax; + function backTrack( + strs: string[], m: number, n: number, + startIndex: number, route: string[] + ): void { + if (startIndex === strs.length) return; + for (let i = startIndex, length = strs.length; i < length; i++) { + route.push(strs[i]); + if (isValidSubSet(route, m, n)) { + resMax = Math.max(resMax, route.length); + backTrack(strs, m, n, i + 1, route); + } + route.pop(); + } + } +}; +function isValidSubSet(strs: string[], m: number, n: number): boolean { + let zeroNum: number = 0, + oneNum: number = 0; + strs.forEach(str => { + for (let s of str) { + if (s === '0') { + zeroNum++; + } else { + oneNum++; + } + } + }); + return zeroNum <= m && oneNum <= n; +} +``` + ----------------------- diff --git a/problems/0494.目标和.md b/problems/0494.目标和.md index 99b76834..8ce1f6f1 100644 --- a/problems/0494.目标和.md +++ b/problems/0494.目标和.md @@ -351,6 +351,25 @@ const findTargetSumWays = (nums, target) => { }; ``` +TypeScript： + +```typescript +function findTargetSumWays(nums: number[], target: number): number { + const sum: number = nums.reduce((pre, cur) => pre + cur); + if (Math.abs(target) > sum) return 0; + if ((target + sum) % 2 === 1) return 0; + const bagSize: number = (target + sum) / 2; + const dp: number[] = new Array(bagSize + 1).fill(0); + dp[0] = 1; + for (let i = 0; i < nums.length; i++) { + for (let j = bagSize; j >= nums[i]; j--) { + dp[j] += dp[j - nums[i]]; + } + } + return dp[bagSize]; +}; +``` + ----------------------- diff --git a/problems/0518.零钱兑换II.md b/problems/0518.零钱兑换II.md index 0e4a3987..b6593438 100644 --- a/problems/0518.零钱兑换II.md +++ b/problems/0518.零钱兑换II.md @@ -274,6 +274,21 @@ const change = (amount, coins) => { } ``` +TypeScript： + +```typescript +function change(amount: number, coins: number[]): number { + const dp: number[] = new Array(amount + 1).fill(0); + dp[0] = 1; + for (let i = 0, length = coins.length; i < length; i++) { + for (let j = coins[i]; j <= amount; j++) { + dp[j] += dp[j - coins[i]]; + } + } + return dp[amount]; +}; +``` + ----------------------- diff --git a/problems/0701.二叉搜索树中的插入操作.md b/problems/0701.二叉搜索树中的插入操作.md index df6a3954..50e39ade 100644 --- a/problems/0701.二叉搜索树中的插入操作.md +++ b/problems/0701.二叉搜索树中的插入操作.md @@ -279,7 +279,7 @@ class Solution: root.right = self.insertIntoBST(root.right, val) # 返回更新后的以当前root为根节点的新树 - return roo + return root ``` **递归法** - 无返回值 diff --git a/problems/0704.二分查找.md b/problems/0704.二分查找.md index 55625130..1e474f9a 100644 --- a/problems/0704.二分查找.md +++ b/problems/0704.二分查找.md @@ -610,7 +610,48 @@ public class Solution{ } } ``` +**Scala:** +（版本一）左闭右闭区间 +```scala +object Solution { + def search(nums: Array[Int], target: Int): Int = { + var left = 0 + var right = nums.length - 1 + while (left <= right) { + var mid = left + ((right - left) / 2) + if (target == nums(mid)) { + return mid + } else if (target < nums(mid)) { + right = mid - 1 + } else { + left = mid + 1 + } + } + -1 + } +} +``` +（版本二）左闭右开区间 +```scala +object Solution { + def search(nums: Array[Int], target: Int): Int = { + var left = 0 + var right = nums.length + while (left < right) { + val mid = left + (right - left) / 2 + if (target == nums(mid)) { + return mid + } else if (target < nums(mid)) { + right = mid + } else { + left = mid + 1 + } + } + -1 + } +} +``` -----------------------
diff --git a/problems/0707.设计链表.md b/problems/0707.设计链表.md index 37ce15ad..dcdb53f4 100644 --- a/problems/0707.设计链表.md +++ b/problems/0707.设计链表.md @@ -1154,7 +1154,75 @@ class MyLinkedList { } ``` +Scala: +```scala +class ListNode(_x: Int = 0, _next: ListNode = null) { + var next: ListNode = _next + var x: Int = _x +} +class MyLinkedList() { + + var size = 0 // 链表尺寸 + var dummy: ListNode = new ListNode(0) // 虚拟头节点 + + // 获取第index个节点的值 + def get(index: Int): Int = { + if (index < 0 || index >= size) { + return -1; + } + var cur = dummy + for (i <- 0 to index) { + cur = cur.next + } + cur.x // 返回cur的值 + } + + // 在链表最前面插入一个节点 + def addAtHead(`val`: Int) { + addAtIndex(0, `val`) + } + + // 在链表最后面插入一个节点 + def addAtTail(`val`: Int) { + addAtIndex(size, `val`) + } + + // 在第index个节点之前插入一个新节点 + // 如果index等于链表长度，则说明新插入的节点是尾巴 + // 如果index等于0，则说明新插入的节点是头 + // 如果index>链表长度，则说明为空 + def addAtIndex(index: Int, `val`: Int) { + if (index > size) { + return + } + var loc = index // 因为参数index是val不可变类型，所以需要赋值给一个可变类型 + if (index < 0) { + loc = 0 + } + size += 1 //链表尺寸+1 + var pre = dummy + for (i <- 0 until loc) { + pre = pre.next + } + val node: ListNode = new ListNode(`val`, pre.next) + pre.next = node + } + // 删除第index个节点 + def deleteAtIndex(index: Int) { + if (index < 0 || index >= size) { + return + } + size -= 1 + var pre = dummy + for (i <- 0 until index) { + pre = pre.next + } + pre.next = pre.next.next + } + +} +``` ----------------------- diff --git a/problems/0977.有序数组的平方.md b/problems/0977.有序数组的平方.md index 24276bcf..0e79a3d6 100644 --- a/problems/0977.有序数组的平方.md +++ b/problems/0977.有序数组的平方.md @@ -358,7 +358,41 @@ class Solution { } } ``` +Scala: +双指针: +```scala +object Solution { + def sortedSquares(nums: Array[Int]): Array[Int] = { + val res: Array[Int] = new Array[Int](nums.length) + var top = nums.length - 1 + var i = 0 + var j = nums.length - 1 + while (i <= j) { + if (nums(i) * nums(i) <= nums(j) * nums(j)) { + // 当左侧平方小于等于右侧，res数组顶部放右侧的平方，并且top下移，j左移 + res(top) = nums(j) * nums(j) + top -= 1 + j -= 1 + } else { + // 当左侧平方大于右侧，res数组顶部放左侧的平方，并且top下移，i右移 + res(top) = nums(i) * nums(i) + top -= 1 + i += 1 + } + } + res + } +} +``` +骚操作(暴力思路): +```scala +object Solution { + def sortedSquares(nums: Array[Int]): Array[Int] = { + nums.map(x=>{x*x}).sortWith(_ < _) + } +} +``` ----------------------- diff --git a/problems/1002.查找常用字符.md b/problems/1002.查找常用字符.md index 36937b0b..075b5ef1 100644 --- a/problems/1002.查找常用字符.md +++ b/problems/1002.查找常用字符.md @@ -418,6 +418,38 @@ char ** commonChars(char ** words, int wordsSize, int* returnSize){ return ret; } ``` - +Scala: +```scala +object Solution { + def commonChars(words: Array[String]): List[String] = { + // 声明返回结果的不可变List集合，因为res要重新赋值，所以声明为var + var res = List[String]() + var hash = new Array[Int](26) // 统计字符出现的最小频率 + // 统计第一个字符串中字符出现的次数 + for (i <- 0 until words(0).length) { + hash(words(0)(i) - 'a') += 1 + } + // 统计其他字符串出现的频率 + for (i <- 1 until words.length) { + // 统计其他字符出现的频率 + var hashOtherStr = new Array[Int](26) + for (j <- 0 until words(i).length) { + hashOtherStr(words(i)(j) - 'a') += 1 + } + // 更新hash，取26个字母最小出现的频率 + for (k <- 0 until 26) { + hash(k) = math.min(hash(k), hashOtherStr(k)) + } + } + // 根据hash的结果转换输出的形式 + for (i <- 0 until 26) { + for (j <- 0 until hash(i)) { + res = res :+ (i + 'a').toChar.toString + } + } + res + } +} +``` -----------------------
diff --git a/problems/1005.K次取反后最大化的数组和.md b/problems/1005.K次取反后最大化的数组和.md index 79767deb..202534da 100644 --- a/problems/1005.K次取反后最大化的数组和.md +++ b/problems/1005.K次取反后最大化的数组和.md @@ -209,6 +209,22 @@ var largestSumAfterKNegations = function(nums, k) { return a + b }) }; + +// 版本二 (优化: 一次遍历) +var largestSumAfterKNegations = function(nums, k) { + nums.sort((a, b) => Math.abs(b) - Math.abs(a)); // 排序 + let sum = 0; + for(let i = 0; i < nums.length; i++) { + if(nums[i] < 0 && k-- > 0) { // 负数取反（k 数量足够时） + nums[i] = -nums[i]; + } + sum += nums[i]; // 求和 + } + if(k % 2 > 0) { // k 有多余的（k若消耗完则应为 -1） + sum -= 2 * nums[nums.length - 1]; // 减去两倍的最小值（因为之前加过一次） + } + return sum; +}; ``` diff --git a/problems/1049.最后一块石头的重量II.md b/problems/1049.最后一块石头的重量II.md index ee0ddef2..3d256c3d 100644 --- a/problems/1049.最后一块石头的重量II.md +++ b/problems/1049.最后一块石头的重量II.md @@ -277,5 +277,26 @@ var lastStoneWeightII = function (stones) { }; ``` +TypeScript： + +```typescript +function lastStoneWeightII(stones: number[]): number { + const sum: number = stones.reduce((pre, cur) => pre + cur); + const bagSize: number = Math.floor(sum / 2); + const weightArr: number[] = stones; + const valueArr: number[] = stones; + const goodsNum: number = weightArr.length; + const dp: number[] = new Array(bagSize + 1).fill(0); + for (let i = 0; i < goodsNum; i++) { + for (let j = bagSize; j >= weightArr[i]; j--) { + dp[j] = Math.max(dp[j], dp[j - weightArr[i]] + valueArr[i]); + } + } + return sum - dp[bagSize] * 2; +}; +``` + + + -----------------------
diff --git a/problems/背包理论基础01背包-1.md b/problems/背包理论基础01背包-1.md index d6bc5520..a40a92ab 100644 --- a/problems/背包理论基础01背包-1.md +++ b/problems/背包理论基础01背包-1.md @@ -380,57 +380,125 @@ func main() { ### javascript ```js -/** - * - * @param {Number []} weight - * @param {Number []} value - * @param {Number} size - * @returns - */ +function testWeightBagProblem (weight, value, size) { + // 定义 dp 数组 + const len = weight.length, + dp = Array(len).fill().map(() => Array(size + 1).fill(0)); -function testWeightBagProblem(weight, value, size) { -const len = weight.length, -dp = Array.from({length: len}).map( -() => Array(size + 1)) //JavaScript 数组是引用类型 -for(let i = 0; i < len; i++) { //初始化最左一列，即背包容量为0时的情况 -dp[i][0] = 0; -} -for(let j = 1; j < size+1; j++) { //初始化第0行, 只有一件物品的情况 -if(weight[0] <= j) { -dp[0][j] = value[0]; -} else { -dp[0][j] = 0; -} -} - -for(let i = 1; i < len; i++) { //dp[i][j]由其左上方元素推导得出 -for(let j = 1; j < size+1; j++) { -if(j < weight[i]) dp[i][j] = dp[i - 1][j]; -else dp[i][j] = Math.max(dp[i-1][j], dp[i-1][j - weight[i]] + value[i]); -} -} - -return dp[len-1][size] //满足条件的最大值 -} - -function testWeightBagProblem2 (wight, value, size) { - const len = wight.length, - dp = Array(size + 1).fill(0); - for(let i = 1; i <= len; i++) { - for(let j = size; j >= wight[i - 1]; j--) { - dp[j] = Math.max(dp[j], value[i - 1] + dp[j - wight[i - 1]]); + // 初始化 + for(let j = weight[0]; j <= size; j++) { + dp[0][j] = value[0]; } - } - return dp[size]; -} + // weight 数组的长度len 就是物品个数 + for(let i = 1; i < len; i++) { // 遍历物品 + for(let j = 0; j <= size; j++) { // 遍历背包容量 + if(j < weight[i]) dp[i][j] = dp[i - 1][j]; + else dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - weight[i]] + value[i]); + } + } + + console.table(dp) + + return dp[len - 1][size]; +} function test () { - console.log(testWeightBagProblem([1, 3, 4, 5], [15, 20, 30, 55], 6)); + console.log(testWeightBagProblem([1, 3, 4, 5], [15, 20, 30, 55], 6)); } test(); ``` + +### C +```c +#include +#include +#include + +#define MAX(a, b) (((a) > (b)) ? (a) : (b)) +#define ARR_SIZE(a) (sizeof((a)) / sizeof((a)[0])) +#define BAG_WEIGHT 4 + +void backPack(int* weights, int weightSize, int* costs, int costSize, int bagWeight) { + // 开辟dp数组 + int dp[weightSize][bagWeight + 1]; + memset(dp, 0, sizeof(int) * weightSize * (bagWeight + 1)); + + int i, j; + // 当背包容量大于物品0的重量时，将物品0放入到背包中 + for(j = weights[0]; j <= bagWeight; ++j) { + dp[0][j] = costs[0]; + } + + // 先遍历物品，再遍历重量 + for(j = 1; j <= bagWeight; ++j) { + for(i = 1; i < weightSize; ++i) { + // 如果当前背包容量小于物品重量 + if(j < weights[i]) + // 背包物品的价值等于背包不放置当前物品时的价值 + dp[i][j] = dp[i-1][j]; + // 若背包当前重量可以放置物品 + else + // 背包的价值等于放置该物品或不放置该物品的最大值 + dp[i][j] = MAX(dp[i - 1][j], dp[i - 1][j - weights[i]] + costs[i]); + } + } + + printf("%d\n", dp[weightSize - 1][bagWeight]); +} + +int main(int argc, char* argv[]) { + int weights[] = {1, 3, 4}; + int costs[] = {15, 20, 30}; + backPack(weights, ARR_SIZE(weights), costs, ARR_SIZE(costs), BAG_WEIGHT); + return 0; +} +``` + + +### TypeScript + +```typescript +function testWeightBagProblem( + weight: number[], + value: number[], + size: number +): number { + /** + * dp[i][j]: 前i个物品，背包容量为j，能获得的最大价值 + * dp[0][*]: u=weight[0],u之前为0，u之后（含u）为value[0] + * dp[*][0]: 0 + * ... + * dp[i][j]: max(dp[i-1][j], dp[i-1][j-weight[i]]+value[i]); + */ + const goodsNum: number = weight.length; + const dp: number[][] = new Array(goodsNum) + .fill(0) + .map((_) => new Array(size + 1).fill(0)); + for (let i = weight[0]; i <= size; i++) { + dp[0][i] = value[0]; + } + for (let i = 1; i < goodsNum; i++) { + for (let j = 1; j <= size; j++) { + if (j < weight[i]) { + dp[i][j] = dp[i - 1][j]; + } else { + dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - weight[i]] + value[i]); + } + } + } + return dp[goodsNum - 1][size]; +} +// test +const weight = [1, 3, 4]; +const value = [15, 20, 30]; +const size = 4; +console.log(testWeightBagProblem(weight, value, size)); +``` + + + -----------------------
diff --git a/problems/背包理论基础01背包-2.md b/problems/背包理论基础01背包-2.md index dabdfb2d..b66b74a6 100644 --- a/problems/背包理论基础01背包-2.md +++ b/problems/背包理论基础01背包-2.md @@ -137,6 +137,8 @@ dp[1] = dp[1 - weight[0]] + value[0] = 15 因为一维dp的写法，背包容量一定是要倒序遍历（原因上面已经讲了），如果遍历背包容量放在上一层，那么每个dp[j]就只会放入一个物品，即：背包里只放入了一个物品。 +倒序遍历的原因是，本质上还是一个对二维数组的遍历，并且右下角的值依赖上一层左上角的值，因此需要保证左边的值仍然是上一层的，从右向左覆盖。 + （这里如果读不懂，就在回想一下dp[j]的定义，或者就把两个for循环顺序颠倒一下试试！） **所以一维dp数组的背包在遍历顺序上和二维其实是有很大差异的！**，这一点大家一定要注意。 @@ -315,6 +317,64 @@ function test () { test(); ``` +### C +```c +#include +#include + +#define MAX(a, b) (((a) > (b)) ? (a) : (b)) +#define ARR_SIZE(arr) ((sizeof((arr))) / sizeof((arr)[0])) +#define BAG_WEIGHT 4 + +void test_back_pack(int* weights, int weightSize, int* values, int valueSize, int bagWeight) { + int dp[bagWeight + 1]; + memset(dp, 0, sizeof(int) * (bagWeight + 1)); + + int i, j; + // 先遍历物品 + for(i = 0; i < weightSize; ++i) { + // 后遍历重量。从后向前遍历 + for(j = bagWeight; j >= weights[i]; --j) { + dp[j] = MAX(dp[j], dp[j - weights[i]] + values[i]); + } + } + + // 打印最优结果 + printf("%d\n", dp[bagWeight]); +} + +int main(int argc, char** argv) { + int weights[] = {1, 3, 4}; + int values[] = {15, 20, 30}; + test_back_pack(weights, ARR_SIZE(weights), values, ARR_SIZE(values), BAG_WEIGHT); + return 0; +} +``` + +### TypeScript + +```typescript +function testWeightBagProblem( + weight: number[], + value: number[], + size: number +): number { + const goodsNum: number = weight.length; + const dp: number[] = new Array(size + 1).fill(0); + for (let i = 0; i < goodsNum; i++) { + for (let j = size; j >= weight[i]; j--) { + dp[j] = Math.max(dp[j], dp[j - weight[i]] + value[i]); + } + } + return dp[size]; +} +const weight = [1, 3, 4]; +const value = [15, 20, 30]; +const size = 4; +console.log(testWeightBagProblem(weight, value, size)); + +``` + ----------------------- diff --git a/problems/背包问题理论基础多重背包.md b/problems/背包问题理论基础多重背包.md index a988db2c..712380f4 100644 --- a/problems/背包问题理论基础多重背包.md +++ b/problems/背包问题理论基础多重背包.md @@ -334,6 +334,64 @@ func Test_multiplePack(t *testing.T) { PASS ``` +TypeScript： + +> 版本一（改变数据源）： + +```typescript +function testMultiPack() { + const bagSize: number = 10; + const weightArr: number[] = [1, 3, 4], + valueArr: number[] = [15, 20, 30], + amountArr: number[] = [2, 3, 2]; + for (let i = 0, length = amountArr.length; i < length; i++) { + while (amountArr[i] > 1) { + weightArr.push(weightArr[i]); + valueArr.push(valueArr[i]); + amountArr[i]--; + } + } + const goodsNum: number = weightArr.length; + const dp: number[] = new Array(bagSize + 1).fill(0); + // 遍历物品 + for (let i = 0; i < goodsNum; i++) { + // 遍历背包容量 + for (let j = bagSize; j >= weightArr[i]; j--) { + dp[j] = Math.max(dp[j], dp[j - weightArr[i]] + valueArr[i]); + } + } + console.log(dp); +} +testMultiPack(); +``` + +> 版本二（改变遍历方式）： + +```typescript +function testMultiPack() { + const bagSize: number = 10; + const weightArr: number[] = [1, 3, 4], + valueArr: number[] = [15, 20, 30], + amountArr: number[] = [2, 3, 2]; + const goodsNum: number = weightArr.length; + const dp: number[] = new Array(bagSize + 1).fill(0); + // 遍历物品 + for (let i = 0; i < goodsNum; i++) { + // 遍历物品个数 + for (let j = 0; j < amountArr[i]; j++) { + // 遍历背包容量 + for (let k = bagSize; k >= weightArr[i]; k--) { + dp[k] = Math.max(dp[k], dp[k - weightArr[i]] + valueArr[i]); + } + } + } + console.log(dp); +} +testMultiPack(); +``` + + + -----------------------
diff --git a/problems/背包问题理论基础完全背包.md b/problems/背包问题理论基础完全背包.md index 3ec399f1..54e772e0 100644 --- a/problems/背包问题理论基础完全背包.md +++ b/problems/背包问题理论基础完全背包.md @@ -340,6 +340,27 @@ function test_completePack2() { } ``` +TypeScript： + +```typescript +// 先遍历物品，再遍历背包容量 +function test_CompletePack(): void { + const weight: number[] = [1, 3, 4]; + const value: number[] = [15, 20, 30]; + const bagSize: number = 4; + const dp: number[] = new Array(bagSize + 1).fill(0); + for (let i = 0; i < weight.length; i++) { + for (let j = weight[i]; j <= bagSize; j++) { + dp[j] = Math.max(dp[j], dp[j - weight[i]] + value[i]); + } + } + console.log(dp); +} +test_CompletePack(); +``` + + + -----------------------
diff --git a/problems/链表理论基础.md b/problems/链表理论基础.md index 2fe9f14c..1a29c32a 100644 --- a/problems/链表理论基础.md +++ b/problems/链表理论基础.md @@ -210,6 +210,13 @@ type ListNode struct { } ``` +Scala: +```scala +class ListNode(_x: Int = 0, _next: ListNode = null) { + var next: ListNode = _next + var x: Int = _x +} +``` ----------------------- diff --git a/problems/面试题 02.07. 解法更新.md b/problems/面试题 02.07. 解法更新.md deleted file mode 100644 index 6115d02e..00000000 --- a/problems/面试题 02.07. 解法更新.md +++ /dev/null @@ -1,41 +0,0 @@ -# 双指针，不计算链表长度 -设置指向headA和headB的指针pa、pb，分别遍历两个链表，每次循环同时更新pa和pb。 -* 当链表A遍历完之后，即pa为空时，将pa指向headB； -* 当链表B遍历完之后，即pa为空时，将pb指向headA； -* 当pa与pb相等时，即指向同一个节点，该节点即为相交起始节点。 -* 若链表不相交，则pa、pb同时为空时退出循环，即如果链表不相交，pa与pb在遍历过全部节点后同时指向结尾空节点，此时退出循环，返回空。 -# 证明思路 -设链表A不相交部分长度为a，链表B不相交部分长度为b，两个链表相交部分长度为c。
-在pa指向链表A时，即pa为空之前，pa经过链表A不相交部分和相交部分，走过的长度为a+c；
-pa指向链表B后，在移动相交节点之前经过链表B不相交部分，走过的长度为b，总合为a+c+b。
-同理，pb走过长度的总合为b+c+a。二者相等，即pa与pb可同时到达相交起始节点。
-该方法可避免计算具体链表长度。 -```cpp -class Solution { -public: - ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { - //链表为空时，返回空指针 - if(headA == nullptr || headB == nullptr) return nullptr; - ListNode* pa = headA; - ListNode* pb = headB; - //pa与pb在遍历过全部节点后,同时指向结尾空节点时退出循环 - while(pa != nullptr || pb != nullptr){ - //pa为空时，将pa指向headB - if(pa == nullptr){ - pa = headB; - } - //pa为空时，将pb指向headA - if(pb == nullptr){ - pb = headA; - } - //pa与pb相等时，返回相交起始节点 - if(pa == pb){ - return pa; - } - pa = pa->next; - pb = pb->next; - } - return nullptr; - } -}; -``` diff --git a/problems/面试题02.07.链表相交.md b/problems/面试题02.07.链表相交.md index 2e7226de..0a38cc33 100644 --- a/problems/面试题02.07.链表相交.md +++ b/problems/面试题02.07.链表相交.md @@ -317,7 +317,55 @@ ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { } ``` - +Scala: +```scala +object Solution { + def getIntersectionNode(headA: ListNode, headB: ListNode): ListNode = { + var lenA = 0 // headA链表的长度 + var lenB = 0 // headB链表的长度 + var tmp = headA // 临时变量 + // 统计headA的长度 + while (tmp != null) { + lenA += 1; + tmp = tmp.next + } + // 统计headB的长度 + tmp = headB // 临时变量赋值给headB + while (tmp != null) { + lenB += 1 + tmp = tmp.next + } + // 因为传递过来的参数是不可变量，所以需要重新定义 + var listA = headA + var listB = headB + // 两个链表的长度差 + // 如果gap>0，lenA>lenB，headA(listA)链表往前移动gap步 + // 如果gap<0，lenA 0) { + // 因为不可以i-=1，所以可以使用for + for (i <- 0 until gap) { + listA = listA.next // 链表headA(listA) 移动 + } + } else { + gap = math.abs(gap) // 此刻gap为负值，取绝对值 + for (i <- 0 until gap) { + listB = listB.next + } + } + // 现在两个链表同时往前走，如果相等则返回 + while (listA != null && listB != null) { + if (listA == listB) { + return listA + } + listA = listA.next + listB = listB.next + } + // 如果链表没有相交则返回null，return可以省略 + null + } +} +``` -----------------------