diff --git a/problems/0045.跳跃游戏II.md b/problems/0045.跳跃游戏II.md index 7a3f048c..4caff042 100644 --- a/problems/0045.跳跃游戏II.md +++ b/problems/0045.跳跃游戏II.md @@ -250,6 +250,27 @@ var jump = function(nums) { }; ``` +### TypeScript + +```typescript +function jump(nums: number[]): number { + const length: number = nums.length; + let curFarthestIndex: number = 0, + nextFarthestIndex: number = 0; + let curIndex: number = 0; + let stepNum: number = 0; + while (curIndex < length - 1) { + nextFarthestIndex = Math.max(nextFarthestIndex, curIndex + nums[curIndex]); + if (curIndex === curFarthestIndex) { + curFarthestIndex = nextFarthestIndex; + stepNum++; + } + curIndex++; + } + return stepNum; +}; +``` + diff --git a/problems/0055.跳跃游戏.md b/problems/0055.跳跃游戏.md index c0890f75..94614242 100644 --- a/problems/0055.跳跃游戏.md +++ b/problems/0055.跳跃游戏.md @@ -154,6 +154,23 @@ var canJump = function(nums) { }; ``` +### TypeScript + +```typescript +function canJump(nums: number[]): boolean { + let farthestIndex: number = 0; + let cur: number = 0; + while (cur <= farthestIndex) { + farthestIndex = Math.max(farthestIndex, cur + nums[cur]); + if (farthestIndex >= nums.length - 1) return true; + cur++; + } + return false; +}; +``` + + + -----------------------
diff --git a/problems/0122.买卖股票的最佳时机II.md b/problems/0122.买卖股票的最佳时机II.md index b31cbae3..4f585a3c 100644 --- a/problems/0122.买卖股票的最佳时机II.md +++ b/problems/0122.买卖股票的最佳时机II.md @@ -268,6 +268,18 @@ const maxProfit = (prices) => { }; ``` +TypeScript： + +```typescript +function maxProfit(prices: number[]): number { + let resProfit: number = 0; + for (let i = 1, length = prices.length; i < length; i++) { + resProfit += Math.max(prices[i] - prices[i - 1], 0); + } + return resProfit; +}; +``` + C: ```c diff --git a/problems/0707.设计链表.md b/problems/0707.设计链表.md index 86f3a683..37ce15ad 100644 --- a/problems/0707.设计链表.md +++ b/problems/0707.设计链表.md @@ -973,6 +973,10 @@ class MyLinkedList { // 处理头节点 if (index === 0) { this.head = this.head!.next; + // 如果链表中只有一个元素，删除头节点后，需要处理尾节点 + if (index === this.size - 1) { + this.tail = null + } this.size--; return; } diff --git a/problems/0763.划分字母区间.md b/problems/0763.划分字母区间.md index 03d3a73b..d350f255 100644 --- a/problems/0763.划分字母区间.md +++ b/problems/0763.划分字母区间.md @@ -77,6 +77,61 @@ public: 但这道题目的思路是很巧妙的，所以有必要介绍给大家做一做，感受一下。 +## 补充 + +这里提供一种与[452.用最少数量的箭引爆气球](https://programmercarl.com/0452.用最少数量的箭引爆气球.html)、[435.无重叠区间](https://programmercarl.com/0435.无重叠区间.html)相同的思路。 + +统计字符串中所有字符的起始和结束位置，记录这些区间(实际上也就是[435.无重叠区间](https://programmercarl.com/0435.无重叠区间.html)题目里的输入)，**将区间按左边界从小到大排序，找到边界将区间划分成组，互不重叠。找到的边界就是答案。** + +```CPP +class Solution { +public: + static bool cmp(vector &a, vector &b) { + return a[0] < b[0]; + } + // 记录每个字母出现的区间 + vector> countLabels(string s) { + vector> hash(26, vector(2, INT_MIN)); + vector> hash_filter; + for (int i = 0; i < s.size(); ++i) { + if (hash[s[i] - 'a'][0] == INT_MIN) { + hash[s[i] - 'a'][0] = i; + } + hash[s[i] - 'a'][1] = i; + } + // 去除字符串中未出现的字母所占用区间 + for (int i = 0; i < hash.size(); ++i) { + if (hash[i][0] != INT_MIN) { + hash_filter.push_back(hash[i]); + } + } + return hash_filter; + } + vector partitionLabels(string s) { + vector res; + // 这一步得到的 hash 即为无重叠区间题意中的输入样例格式：区间列表 + // 只不过现在我们要求的是区间分割点 + vector> hash = countLabels(s); + // 按照左边界从小到大排序 + sort(hash.begin(), hash.end(), cmp); + // 记录最大右边界 + int rightBoard = hash[0][1]; + int leftBoard = 0; + for (int i = 1; i < hash.size(); ++i) { + // 由于字符串一定能分割，因此, + // 一旦下一区间左边界大于当前右边界，即可认为出现分割点 + if (hash[i][0] > rightBoard) { + res.push_back(rightBoard - leftBoard + 1); + leftBoard = hash[i][0]; + } + rightBoard = max(rightBoard, hash[i][1]); + } + // 最右端 + res.push_back(rightBoard - leftBoard + 1); + return res; + } +}; +``` ## 其他语言版本 diff --git a/problems/前序/ACM模式如何构建二叉树.md b/problems/前序/ACM模式如何构建二叉树.md index 444f0071..815eb17d 100644 --- a/problems/前序/ACM模式如何构建二叉树.md +++ b/problems/前序/ACM模式如何构建二叉树.md @@ -208,6 +208,59 @@ int main() { ## Java ```Java +public class Solution { + // 节点类 + static class TreeNode { + // 节点值 + int val; + + // 左节点 + TreeNode left; + + // 右节点 + TreeNode right; + + // 节点的构造函数(默认左右节点都为null) + public TreeNode(int x) { + this.val = x; + this.left = null; + this.right = null; + } + } + + /** + * 根据数组构建二叉树 + * @param arr 树的数组表示 + * @return 构建成功后树的根节点 + */ + public TreeNode constructBinaryTree(final int[] arr) { + // 构建和原数组相同的树节点列表 + List treeNodeList = arr.length > 0 ? new ArrayList<>(arr.length) : null; + TreeNode root = null; + // 把输入数值数组，先转化为二叉树节点列表 + for (int i = 0; i < arr.length; i++) { + TreeNode node = null; + if (arr[i] != -1) { // 用 -1 表示null + node = new TreeNode(arr[i]); + } + treeNodeList.add(node); + if (i == 0) { + root = node; + } + } + // 遍历一遍，根据规则左右孩子赋值就可以了 + // 注意这里 结束规则是 i * 2 + 2 < arr.length，避免空指针 + for (int i = 0; i * 2 + 2 < arr.length; i++) { + TreeNode node = treeNodeList.get(i); + if (node != null) { + // 线性存储转连式存储关键逻辑 + node.left = treeNodeList.get(2 * i + 1); + node.right = treeNodeList.get(2 * i + 2); + } + } + return root; + } +} ```