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https://github.com/youngyangyang04/leetcode-master.git
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更新 0416.分割等和自己 0474.一和零 0494.目标和 1049.最后一块石头的重量II 排版格式修复
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@ -30,9 +30,9 @@
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* 1 <= nums.length <= 200
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* 1 <= nums[i] <= 100
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# 算法公开课
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## 算法公开课
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**《代码随想录》算法视频公开课:[动态规划之背包问题,这个包能装满吗?| LeetCode:416.分割等和子集](https://www.bilibili.com/video/BV1rt4y1N7jE/),相信结合视频再看本篇题解,更有助于大家对本题的理解**。
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**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html):[动态规划之背包问题,这个包能装满吗?| LeetCode:416.分割等和子集](https://www.bilibili.com/video/BV1rt4y1N7jE/),相信结合视频再看本篇题解,更有助于大家对本题的理解**。
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## 思路
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@ -53,7 +53,7 @@
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* [动态规划:关于01背包问题,你该了解这些!](https://programmercarl.com/背包理论基础01背包-1.html)
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* [动态规划:关于01背包问题,你该了解这些!(滚动数组)](https://programmercarl.com/背包理论基础01背包-2.html)
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## 01背包问题
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### 01背包问题
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背包问题,大家都知道,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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@ -479,7 +479,7 @@ func canPartition(nums []int) bool {
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}
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```
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### javaScript:
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### JavaScript:
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```js
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var canPartition = function(nums) {
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@ -499,7 +499,7 @@ var canPartition = function(nums) {
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```
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### Rust
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### Rust:
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```Rust
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impl Solution {
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@ -681,7 +681,7 @@ function canPartition(nums: number[]): boolean {
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};
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```
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### Scala
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### Scala:
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滚动数组:
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```scala
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@ -730,3 +730,4 @@ object Solution {
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<a href="https://programmercarl.com/other/kstar.html" target="_blank">
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<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
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</a>
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@ -34,9 +34,9 @@
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* strs[i] 仅由 '0' 和 '1' 组成
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* 1 <= m, n <= 100
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# 算法公开课
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## 算法公开课
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**《代码随想录》算法视频公开课:[装满这个背包最多用多少个物品?| LeetCode:474.一和零](https://www.bilibili.com/video/BV1rW4y1x7ZQ/),相信结合视频再看本篇题解,更有助于大家对本题的理解**。
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**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html):[装满这个背包最多用多少个物品?| LeetCode:474.一和零](https://www.bilibili.com/video/BV1rW4y1x7ZQ/),相信结合视频再看本篇题解,更有助于大家对本题的理解**。
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## 思路
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@ -538,3 +538,4 @@ impl Solution {
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<a href="https://programmercarl.com/other/kstar.html" target="_blank">
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<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
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</a>
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@ -37,9 +37,9 @@
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* 初始的数组的和不会超过 1000 。
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* 保证返回的最终结果能被 32 位整数存下。
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# 算法公开课
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## 算法公开课
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**《代码随想录》算法视频公开课:[装满背包有多少种方法?| LeetCode:494.目标和](https://www.bilibili.com/video/BV1o8411j73x/),相信结合视频再看本篇题解,更有助于大家对本题的理解**。
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**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html):[装满背包有多少种方法?| LeetCode:494.目标和](https://www.bilibili.com/video/BV1o8411j73x/),相信结合视频再看本篇题解,更有助于大家对本题的理解**。
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## 思路
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@ -67,7 +67,7 @@ target是固定的,sum是固定的,left就可以求出来。
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此时问题就是在集合nums中找出和为left的组合。
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## 回溯算法
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### 回溯算法
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在回溯算法系列中,一起学过这道题目[回溯算法:39. 组合总和](https://programmercarl.com/0039.组合总和.html)的录友应该感觉很熟悉,这不就是组合总和问题么?
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@ -118,7 +118,7 @@ public:
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也可以使用记忆化回溯,但这里我就不在回溯上下功夫了,直接看动规吧
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## 动态规划
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### 动态规划
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如何转化为01背包问题呢。
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@ -519,8 +519,6 @@ const findTargetSumWays = (nums, target) => {
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### TypeScript
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TypeScript:
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```ts
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function findTargetSumWays(nums: number[], target: number): number {
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// 把数组分成两个组合left, right.left + right = sum, left - right = target.
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@ -590,3 +588,4 @@ impl Solution {
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<a href="https://programmercarl.com/other/kstar.html" target="_blank">
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<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
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</a>
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@ -35,9 +35,9 @@
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* 1 <= stones.length <= 30
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* 1 <= stones[i] <= 1000
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# 算法公开课
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## 算法公开课
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**《代码随想录》算法视频公开课:[这个背包最多能装多少?LeetCode:1049.最后一块石头的重量II](https://www.bilibili.com/video/BV14M411C7oV/),相信结合视频再看本篇题解,更有助于大家对本题的理解**。
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**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html):[这个背包最多能装多少?LeetCode:1049.最后一块石头的重量II](https://www.bilibili.com/video/BV14M411C7oV/),相信结合视频再看本篇题解,更有助于大家对本题的理解**。
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## 思路
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@ -341,7 +341,7 @@ func max(a, b 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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/**
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@ -364,7 +364,7 @@ var lastStoneWeightII = function (stones) {
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};
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```
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### C
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### C:
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```c
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#define MAX(a, b) (((a) > (b)) ? (a) : (b))
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@ -413,7 +413,7 @@ function lastStoneWeightII(stones: number[]): number {
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};
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```
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### Scala
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### Scala:
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滚动数组:
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```scala
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@ -455,7 +455,7 @@ object Solution {
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}
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```
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### Rust
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### Rust:
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```rust
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impl Solution {
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@ -477,3 +477,4 @@ impl Solution {
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<a href="https://programmercarl.com/other/kstar.html" target="_blank">
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<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
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</a>
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