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Update
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youngyangyang04
@ -4,7 +4,7 @@
|
||||
</a>
|
||||
<p align="center"><strong><a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！</strong></p>
|
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## 63. 不同路径 II
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# 63. 不同路径 II
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[力扣题目链接](https://leetcode-cn.com/problems/unique-paths-ii/)
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@ -22,11 +22,11 @@
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输入：obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]]
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输出：2
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* 输入：obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]]
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* 输出：2
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解释：
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3x3 网格的正中间有一个障碍物。
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从左上角到右下角一共有 2 条不同的路径：
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* 3x3 网格的正中间有一个障碍物。
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* 从左上角到右下角一共有 2 条不同的路径：
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1. 向右 -> 向右 -> 向下 -> 向下
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2. 向下 -> 向下 -> 向右 -> 向右
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@ -34,11 +34,10 @@
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输入：obstacleGrid = [[0,1],[0,0]]
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输出：1
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* 输入：obstacleGrid = [[0,1],[0,0]]
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* 输出：1
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提示：
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* m == obstacleGrid.length
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* n == obstacleGrid[i].length
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* 1 <= m, n <= 100
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@ -171,7 +170,7 @@ public:
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## 其他语言版本
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Java：
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### Java
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```java
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class Solution {
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@ -199,7 +198,7 @@ class Solution {
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```
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Python：
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### Python
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```python
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class Solution:
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@ -262,7 +261,7 @@ class Solution:
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```
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Go：
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### Go
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```go
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func uniquePathsWithObstacles(obstacleGrid [][]int) int {
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@ -308,7 +307,7 @@ func uniquePathsWithObstacles(obstacleGrid [][]int) int {
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```
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Javascript
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### Javascript
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```Javascript
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var uniquePathsWithObstacles = function(obstacleGrid) {
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const m = obstacleGrid.length
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@ -4,7 +4,7 @@
|
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</a>
|
||||
<p align="center"><strong><a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！</strong></p>
|
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## 96.不同的二叉搜索树
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# 96.不同的二叉搜索树
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[力扣题目链接](https://leetcode-cn.com/problems/unique-binary-search-trees/)
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@ -163,7 +163,7 @@ public:
|
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## 其他语言版本
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|
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|
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Java：
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### Java
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```Java
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class Solution {
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public int numTrees(int n) {
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@ -184,7 +184,7 @@ class Solution {
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}
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```
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Python：
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### Python
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```python
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class Solution:
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def numTrees(self, n: int) -> int:
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@ -196,7 +196,7 @@ class Solution:
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return dp[-1]
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```
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Go：
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### Go
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```Go
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func numTrees(n int)int{
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dp:=make([]int,n+1)
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@ -210,7 +210,7 @@ func numTrees(n int)int{
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}
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```
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Javascript：
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### Javascript
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```Javascript
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const numTrees =(n) => {
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let dp = new Array(n+1).fill(0);
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@ -4,23 +4,22 @@
|
||||
</a>
|
||||
<p align="center"><strong><a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！</strong></p>
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## 343. 整数拆分
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# 343. 整数拆分
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[力扣题目链接](https://leetcode-cn.com/problems/integer-break/)
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给定一个正整数 n，将其拆分为至少两个正整数的和，并使这些整数的乘积最大化。 返回你可以获得的最大乘积。
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示例 1:
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输入: 2
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输出: 1
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\解释: 2 = 1 + 1, 1 × 1 = 1。
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* 输入: 2
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* 输出: 1
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* 解释: 2 = 1 + 1, 1 × 1 = 1。
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示例 2:
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输入: 10
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输出: 36
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解释: 10 = 3 + 3 + 4, 3 × 3 × 4 = 36。
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说明: 你可以假设 n 不小于 2 且不大于 58。
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* 输入: 10
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* 输出: 36
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* 解释: 10 = 3 + 3 + 4, 3 × 3 × 4 = 36。
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* 说明: 你可以假设 n 不小于 2 且不大于 58。
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## 思路
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@ -193,7 +192,7 @@ public:
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## 其他语言版本
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||||
|
||||
|
||||
Java：
|
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### Java
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||||
```Java
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class Solution {
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public int integerBreak(int n) {
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@ -212,7 +211,7 @@ class Solution {
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||||
}
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||||
```
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Python：
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||||
### Python
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||||
```python
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||||
class Solution:
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def integerBreak(self, n: int) -> int:
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@ -226,7 +225,8 @@ class Solution:
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dp[i] = max(dp[i], max(j * (i - j), j * dp[i - j]))
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return dp[n]
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```
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Go：
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||||
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||||
### Go
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```golang
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func integerBreak(n int) int {
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/**
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@ -256,7 +256,7 @@ func max(a,b int) int{
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||||
}
|
||||
```
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||||
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||||
Javascript:
|
||||
### Javascript
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||||
```Javascript
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var integerBreak = function(n) {
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let dp = new Array(n + 1).fill(0)
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||||
@ -3,11 +3,12 @@
|
||||
<img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20210924105952.png" width="1000"/>
|
||||
</a>
|
||||
<p align="center"><strong><a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！</strong></p>
|
||||
|
||||
# 动态规划：关于01背包问题，你该了解这些！
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||||
这周我们正式开始讲解背包问题！
|
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|
||||
背包问题的经典资料当然是：背包九讲。在公众号「代码随想录」后台回复：背包九讲，就可以获得背包九讲的PDF。
|
||||
背包问题的经典资料当然是：背包九讲。在公众号「代码随想录」后台回复：背包九讲，就可以获得背包九讲的pdf。
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||||
|
||||
但说实话，背包九讲对于小白来说确实不太友好，看起来还是有点费劲的，而且都是伪代码理解起来也吃力。
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@ -32,7 +33,7 @@ leetcode上没有纯01背包的问题，都是01背包应用方面的题目，
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## 01 背包
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有N件物品和一个最多能背重量为W 的背包。第i件物品的重量是weight[i]，得到的价值是value[i] 。**每件物品只能用一次**，求解将哪些物品装入背包里物品价值总和最大。
|
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有n件物品和一个最多能背重量为w 的背包。第i件物品的重量是weight[i]，得到的价值是value[i] 。**每件物品只能用一次**，求解将哪些物品装入背包里物品价值总和最大。
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||||
|
||||
|
||||
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||||
@ -40,7 +41,7 @@ leetcode上没有纯01背包的问题，都是01背包应用方面的题目，
|
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|
||||
这样其实是没有从底向上去思考，而是习惯性想到了背包，那么暴力的解法应该是怎么样的呢？
|
||||
|
||||
每一件物品其实只有两个状态，取或者不取，所以可以使用回溯法搜索出所有的情况，那么时间复杂度就是$O(2^n)$，这里的n表示物品数量。
|
||||
每一件物品其实只有两个状态，取或者不取，所以可以使用回溯法搜索出所有的情况，那么时间复杂度就是$o(2^n)$，这里的n表示物品数量。
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||||
|
||||
**所以暴力的解法是指数级别的时间复杂度。进而才需要动态规划的解法来进行优化！**
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||||
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||||
@ -109,7 +110,7 @@ for (int j = 0 ; j < weight[0]; j++) { // 当然这一步，如果把dp数组
|
||||
dp[0][j] = 0;
|
||||
}
|
||||
// 正序遍历
|
||||
for (int j = weight[0]; j <= bagWeight; j++) {
|
||||
for (int j = weight[0]; j <= bagweight; j++) {
|
||||
dp[0][j] = value[0];
|
||||
}
|
||||
```
|
||||
@ -135,8 +136,8 @@ dp[0][j] 和 dp[i][0] 都已经初始化了，那么其他下标应该初始化
|
||||
|
||||
```
|
||||
// 初始化 dp
|
||||
vector<vector<int>> dp(weight.size(), vector<int>(bagWeight + 1, 0));
|
||||
for (int j = weight[0]; j <= bagWeight; j++) {
|
||||
vector<vector<int>> dp(weight.size(), vector<int>(bagweight + 1, 0));
|
||||
for (int j = weight[0]; j <= bagweight; j++) {
|
||||
dp[0][j] = value[0];
|
||||
}
|
||||
|
||||
@ -160,7 +161,7 @@ for (int j = weight[0]; j <= bagWeight; j++) {
|
||||
```
|
||||
// weight数组的大小 就是物品个数
|
||||
for(int i = 1; i < weight.size(); i++) { // 遍历物品
|
||||
for(int j = 0; j <= bagWeight; j++) { // 遍历背包容量
|
||||
for(int j = 0; j <= bagweight; j++) { // 遍历背包容量
|
||||
if (j < weight[i]) dp[i][j] = dp[i - 1][j];
|
||||
else dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - weight[i]] + value[i]);
|
||||
|
||||
@ -174,7 +175,7 @@ for(int i = 1; i < weight.size(); i++) { // 遍历物品
|
||||
|
||||
```
|
||||
// weight数组的大小 就是物品个数
|
||||
for(int j = 0; j <= bagWeight; j++) { // 遍历背包容量
|
||||
for(int j = 0; j <= bagweight; j++) { // 遍历背包容量
|
||||
for(int i = 1; i < weight.size(); i++) { // 遍历物品
|
||||
if (j < weight[i]) dp[i][j] = dp[i - 1][j];
|
||||
else dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - weight[i]] + value[i]);
|
||||
@ -219,32 +220,32 @@ dp[i-1][j]和dp[i - 1][j - weight[i]] 都在dp[i][j]的左上角方向（包括
|
||||
主要就是自己没有动手推导一下dp数组的演变过程，如果推导明白了，代码写出来就算有问题，只要把dp数组打印出来，对比一下和自己推导的有什么差异，很快就可以发现问题了。
|
||||
|
||||
|
||||
## 完整C++测试代码
|
||||
## 完整c++测试代码
|
||||
|
||||
```CPP
|
||||
```cpp
|
||||
void test_2_wei_bag_problem1() {
|
||||
vector<int> weight = {1, 3, 4};
|
||||
vector<int> value = {15, 20, 30};
|
||||
int bagWeight = 4;
|
||||
int bagweight = 4;
|
||||
|
||||
// 二维数组
|
||||
vector<vector<int>> dp(weight.size(), vector<int>(bagWeight + 1, 0));
|
||||
vector<vector<int>> dp(weight.size(), vector<int>(bagweight + 1, 0));
|
||||
|
||||
// 初始化
|
||||
for (int j = weight[0]; j <= bagWeight; j++) {
|
||||
for (int j = weight[0]; j <= bagweight; j++) {
|
||||
dp[0][j] = value[0];
|
||||
}
|
||||
|
||||
// weight数组的大小 就是物品个数
|
||||
for(int i = 1; i < weight.size(); i++) { // 遍历物品
|
||||
for(int j = 0; j <= bagWeight; j++) { // 遍历背包容量
|
||||
for(int j = 0; j <= bagweight; j++) { // 遍历背包容量
|
||||
if (j < weight[i]) dp[i][j] = dp[i - 1][j];
|
||||
else dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - weight[i]] + value[i]);
|
||||
|
||||
}
|
||||
}
|
||||
|
||||
cout << dp[weight.size() - 1][bagWeight] << endl;
|
||||
cout << dp[weight.size() - 1][bagweight] << endl;
|
||||
}
|
||||
|
||||
int main() {
|
||||
@ -267,48 +268,45 @@ int main() {
|
||||
|
||||
## 其他语言版本
|
||||
|
||||
Java：
|
||||
### java
|
||||
|
||||
```java
|
||||
public static void main(String[] args) {
|
||||
public static void main(string[] args) {
|
||||
int[] weight = {1, 3, 4};
|
||||
int[] value = {15, 20, 30};
|
||||
int bagSize = 4;
|
||||
testWeightBagProblem(weight, value, bagSize);
|
||||
int bagsize = 4;
|
||||
testweightbagproblem(weight, value, bagsize);
|
||||
}
|
||||
|
||||
public static void testWeightBagProblem(int[] weight, int[] value, int bagSize){
|
||||
int wLen = weight.length, value0 = 0;
|
||||
public static void testweightbagproblem(int[] weight, int[] value, int bagsize){
|
||||
int wlen = weight.length, value0 = 0;
|
||||
//定义dp数组：dp[i][j]表示背包容量为j时，前i个物品能获得的最大价值
|
||||
int[][] dp = new int[wLen + 1][bagSize + 1];
|
||||
int[][] dp = new int[wlen + 1][bagsize + 1];
|
||||
//初始化：背包容量为0时，能获得的价值都为0
|
||||
for (int i = 0; i <= wLen; i++){
|
||||
for (int i = 0; i <= wlen; i++){
|
||||
dp[i][0] = value0;
|
||||
}
|
||||
//遍历顺序：先遍历物品，再遍历背包容量
|
||||
for (int i = 1; i <= wLen; i++){
|
||||
for (int j = 1; j <= bagSize; j++){
|
||||
for (int i = 1; i <= wlen; i++){
|
||||
for (int j = 1; j <= bagsize; j++){
|
||||
if (j < weight[i - 1]){
|
||||
dp[i][j] = dp[i - 1][j];
|
||||
}else{
|
||||
dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - weight[i - 1]] + value[i - 1]);
|
||||
dp[i][j] = math.max(dp[i - 1][j], dp[i - 1][j - weight[i - 1]] + value[i - 1]);
|
||||
}
|
||||
}
|
||||
}
|
||||
//打印dp数组
|
||||
for (int i = 0; i <= wLen; i++){
|
||||
for (int j = 0; j <= bagSize; j++){
|
||||
System.out.print(dp[i][j] + " ");
|
||||
for (int i = 0; i <= wlen; i++){
|
||||
for (int j = 0; j <= bagsize; j++){
|
||||
system.out.print(dp[i][j] + " ");
|
||||
}
|
||||
System.out.print("\n");
|
||||
system.out.print("\n");
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
Python：
|
||||
### python
|
||||
```python
|
||||
def test_2_wei_bag_problem1(bag_size, weight, value) -> int:
|
||||
rows, cols = len(weight), bag_size + 1
|
||||
@ -343,26 +341,26 @@ if __name__ == "__main__":
|
||||
```
|
||||
|
||||
|
||||
Go：
|
||||
### go
|
||||
```go
|
||||
func test_2_wei_bag_problem1(weight, value []int, bagWeight int) int {
|
||||
func test_2_wei_bag_problem1(weight, value []int, bagweight int) int {
|
||||
// 定义dp数组
|
||||
dp := make([][]int, len(weight))
|
||||
for i, _ := range dp {
|
||||
dp[i] = make([]int, bagWeight+1)
|
||||
dp[i] = make([]int, bagweight+1)
|
||||
}
|
||||
// 初始化
|
||||
for j := bagWeight; j >= weight[0]; j-- {
|
||||
for j := bagweight; j >= weight[0]; j-- {
|
||||
dp[0][j] = dp[0][j-weight[0]] + value[0]
|
||||
}
|
||||
// 递推公式
|
||||
for i := 1; i < len(weight); i++ {
|
||||
//正序,也可以倒序
|
||||
for j := weight[i];j<= bagWeight ; j++ {
|
||||
for j := weight[i];j<= bagweight ; j++ {
|
||||
dp[i][j] = max(dp[i-1][j], dp[i-1][j-weight[i]]+value[i])
|
||||
}
|
||||
}
|
||||
return dp[len(weight)-1][bagWeight]
|
||||
return dp[len(weight)-1][bagweight]
|
||||
}
|
||||
|
||||
func max(a,b int) int {
|
||||
@ -379,19 +377,19 @@ func main() {
|
||||
}
|
||||
```
|
||||
|
||||
javaScript:
|
||||
### javascript
|
||||
|
||||
```js
|
||||
function testWeightBagProblem (wight, value, size) {
|
||||
function testweightbagproblem (wight, value, size) {
|
||||
const len = wight.length,
|
||||
dp = Array.from({length: len + 1}).map(
|
||||
() => Array(size + 1).fill(0)
|
||||
dp = array.from({length: len + 1}).map(
|
||||
() => array(size + 1).fill(0)
|
||||
);
|
||||
|
||||
for(let i = 1; i <= len; i++) {
|
||||
for(let j = 0; j <= size; j++) {
|
||||
if(wight[i - 1] <= j) {
|
||||
dp[i][j] = Math.max(
|
||||
dp[i][j] = math.max(
|
||||
dp[i - 1][j],
|
||||
value[i - 1] + dp[i - 1][j - wight[i - 1]]
|
||||
)
|
||||
|
||||
Reference in New Issue
Block a user