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youngyangyang04 committed May 20, 2023
commit 16f4a48bd6b0539c60eed95c4f2c54a052685386
2 changes: 1 addition & 1 deletion problems/0053.最大子序和.md
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## 动态规划

当然本题还可以用动态规划来做,当前[「代码随想录」](https://img-blog.csdnimg.cn/20201124161234338.png)主要讲解贪心系列,后续到动态规划系列的时候会详细讲解本题的 dp 方法。
当然本题还可以用动态规划来做,在代码随想录动态规划章节我会详细介绍,如果大家想在想看,可以直接跳转:[动态规划版本详解](https://programmercarl.com/0053.%E6%9C%80%E5%A4%A7%E5%AD%90%E5%BA%8F%E5%92%8C%EF%BC%88%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%EF%BC%89.html#%E6%80%9D%E8%B7%AF)

那么先给出我的 dp 代码如下,有时间的录友可以提前做一做:

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返回你可以从这笔交易中获取的最大利润。如果你不能获取任何利润,返回 0 。

示例 1:
输入:[7,1,5,3,6,4]
输出:5
* 示例 1:
* 输入:[7,1,5,3,6,4]
* 输出:5
解释:在第 2 天(股票价格 = 1)的时候买入,在第 5 天(股票价格 = 6)的时候卖出,最大利润 = 6-1 = 5 。注意利润不能是 7-1 = 6, 因为卖出价格需要大于买入价格;同时,你不能在买入前卖出股票。

示例 2:
输入:prices = [7,6,4,3,1]
输出:0
* 示例 2:
* 输入:prices = [7,6,4,3,1]
* 输出:0
解释:在这种情况下, 没有交易完成, 所以最大利润为 0。

# 算法公开课

**《代码随想录》算法视频公开课:[动态规划之 LeetCode:121.买卖股票的最佳时机1](https://www.bilibili.com/video/BV1Xe4y1u77q),相信结合视频再看本篇题解,更有助于大家对本题的理解**。



## 思路

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### 动态规划

动态规划将在下一个系列详细讲解,本题解先给出我的 C++代码(带详细注释),感兴趣的同学可以自己先学习一下。
动态规划将在下一个系列详细讲解,本题解先给出我的 C++代码(带详细注释),想先学习的话,可以看本篇:[122.买卖股票的最佳时机II(动态规划)](https://programmercarl.com/0122.%E4%B9%B0%E5%8D%96%E8%82%A1%E7%A5%A8%E7%9A%84%E6%9C%80%E4%BD%B3%E6%97%B6%E6%9C%BAII%EF%BC%88%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%EF%BC%89.html#%E6%80%9D%E8%B7%AF)

```CPP
class Solution {
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注意:你不能同时参与多笔交易(你必须在再次购买前出售掉之前的股票)。


示例 1:
输入: [7,1,5,3,6,4]
输出: 7
* 示例 1:
* 输入: [7,1,5,3,6,4]
* 输出: 7
解释: 在第 2 天(股票价格 = 1)的时候买入,在第 3 天(股票价格 = 5)的时候卖出, 这笔交易所能获得利润 = 5-1 = 4。随后,在第 4 天(股票价格 = 3)的时候买入,在第 5 天(股票价格 = 6)的时候卖出, 这笔交易所能获得利润 = 6-3 = 3 。

示例 2:
输入: [1,2,3,4,5]
输出: 4
* 示例 2:
* 输入: [1,2,3,4,5]
* 输出: 4
解释: 在第 1 天(股票价格 = 1)的时候买入,在第 5 天 (股票价格 = 5)的时候卖出, 这笔交易所能获得利润 = 5-1 = 4 。注意你不能在第 1 天和第 2 天接连购买股票,之后再将它们卖出。因为这样属于同时参与了多笔交易,你必须在再次购买前出售掉之前的股票。

示例 3:
输入: [7,6,4,3,1]
输出: 0
* 示例 3:
* 输入: [7,6,4,3,1]
* 输出: 0
解释: 在这种情况下, 没有交易完成, 所以最大利润为 0。

提示:
* 1 <= prices.length <= 3 * 10 ^ 4
* 0 <= prices[i] <= 10 ^ 4

# 算法公开课

**《代码随想录》算法视频公开课:[动态规划,股票问题第二弹 | LeetCode:122.买卖股票的最佳时机II](https://www.bilibili.com/video/BV1D24y1Q7Ls),相信结合视频再看本篇题解,更有助于大家对本题的理解**。


## 思路

本题我们在讲解贪心专题的时候就已经讲解过了[贪心算法:买卖股票的最佳时机II](https://programmercarl.com/0122.买卖股票的最佳时机II.html),只不过没有深入讲解动态规划的解法,那么这次我们再好好分析一下动规的解法。
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注意:你不能同时参与多笔交易(你必须在再次购买前出售掉之前的股票)。

示例 1:
输入:prices = [3,3,5,0,0,3,1,4]
输出:6
* 示例 1:
* 输入:prices = [3,3,5,0,0,3,1,4]
* 输出:6
解释:在第 4 天(股票价格 = 0)的时候买入,在第 6 天(股票价格 = 3)的时候卖出,这笔交易所能获得利润 = 3-0 = 3 。随后,在第 7 天(股票价格 = 1)的时候买入,在第 8 天 (股票价格 = 4)的时候卖出,这笔交易所能获得利润 = 4-1 = 3。

示例 2:
输入:prices = [1,2,3,4,5]
输出:4
* 示例 2:
* 输入:prices = [1,2,3,4,5]
* 输出:4
解释:在第 1 天(股票价格 = 1)的时候买入,在第 5 天 (股票价格 = 5)的时候卖出, 这笔交易所能获得利润 = 5-1 = 4。注意你不能在第 1 天和第 2 天接连购买股票,之后再将它们卖出。因为这样属于同时参与了多笔交易,你必须在再次购买前出售掉之前的股票。

示例 3:
输入:prices = [7,6,4,3,1]
输出:0
* 示例 3:
* 输入:prices = [7,6,4,3,1]
* 输出:0
解释:在这个情况下, 没有交易完成, 所以最大利润为0。

示例 4:
输入:prices = [1]
* 示例 4:
* 输入:prices = [1]
输出:0

提示:

* 1 <= prices.length <= 10^5
* 0 <= prices[i] <= 10^5

# 算法公开课

**《代码随想录》算法视频公开课:[动态规划,股票至多买卖两次,怎么求? | LeetCode:123.买卖股票最佳时机III](https://www.bilibili.com/video/BV1WG411K7AR),相信结合视频再看本篇题解,更有助于大家对本题的理解**。


## 思路


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注意:你不能同时参与多笔交易(你必须在再次购买前出售掉之前的股票)。

示例 1:
输入:k = 2, prices = [2,4,1]
输出:2
* 示例 1:
* 输入:k = 2, prices = [2,4,1]
* 输出:2
解释:在第 1 天 (股票价格 = 2) 的时候买入,在第 2 天 (股票价格 = 4) 的时候卖出,这笔交易所能获得利润 = 4-2 = 2。

示例 2:
输入:k = 2, prices = [3,2,6,5,0,3]
输出:7
* 示例 2:
* 输入:k = 2, prices = [3,2,6,5,0,3]
* 输出:7
解释:在第 2 天 (股票价格 = 2) 的时候买入,在第 3 天 (股票价格 = 6) 的时候卖出, 这笔交易所能获得利润 = 6-2 = 4。随后,在第 5 天 (股票价格 = 0) 的时候买入,在第 6 天 (股票价格 = 3) 的时候卖出, 这笔交易所能获得利润 = 3-0 = 3 。


Expand All @@ -31,6 +31,11 @@
* 0 <= prices.length <= 1000
* 0 <= prices[i] <= 1000

# 算法公开课

**《代码随想录》算法视频公开课:[动态规划来决定最佳时机,至多可以买卖K次!| LeetCode:188.买卖股票最佳时机4](https://www.bilibili.com/video/BV16M411U7XJ),相信结合视频再看本篇题解,更有助于大家对本题的理解**。


## 思路

这道题目可以说是[动态规划:123.买卖股票的最佳时机III](https://programmercarl.com/0123.买卖股票的最佳时机III.html)的进阶版,这里要求至多有k次交易。
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* 0 <= nums.length <= 100
* 0 <= nums[i] <= 400

# 算法公开课

**《代码随想录》算法视频公开课:[动态规划,偷不偷这个房间呢?| LeetCode:198.打家劫舍](https://www.bilibili.com/video/BV1Te411N7SX),相信结合视频再看本篇题解,更有助于大家对本题的理解**。


## 思路

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示例 1:

输入:nums = [2,3,2]
输出:3
解释:你不能先偷窃 1 号房屋(金额 = 2),然后偷窃 3 号房屋(金额 = 2), 因为他们是相邻的。
* 输入:nums = [2,3,2]
* 输出:3
* 解释:你不能先偷窃 1 号房屋(金额 = 2),然后偷窃 3 号房屋(金额 = 2), 因为他们是相邻的。

示例 2:
输入:nums = [1,2,3,1]
输出:4
解释:你可以先偷窃 1 号房屋(金额 = 1),然后偷窃 3 号房屋(金额 = 3)。偷窃到的最高金额 = 1 + 3 = 4 。
* 示例 2:
* 输入:nums = [1,2,3,1]
* 输出:4
* 解释:你可以先偷窃 1 号房屋(金额 = 1),然后偷窃 3 号房屋(金额 = 3)。偷窃到的最高金额 = 1 + 3 = 4 。

示例 3:
输入:nums = [0]
输出:0
* 示例 3:
* 输入:nums = [0]
* 输出:0

提示:
* 1 <= nums.length <= 100
* 0 <= nums[i] <= 1000

# 算法公开课

**《代码随想录》算法视频公开课:[动态规划,房间连成环了那还偷不偷呢?| LeetCode:213.打家劫舍II](https://www.bilibili.com/video/BV1oM411B7xq),相信结合视频再看本篇题解,更有助于大家对本题的理解**。


## 思路

这道题目和[198.打家劫舍](https://programmercarl.com/0198.打家劫舍.html)是差不多的,唯一区别就是成环了。
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* 输出: 3
* 解释: 对应的交易状态为: [买入, 卖出, 冷冻期, 买入, 卖出]

# 算法公开课

**《代码随想录》算法视频公开课:[动态规划来决定最佳时机,这次有冷冻期!| LeetCode:309.买卖股票的最佳时机含冷冻期](https://www.bilibili.com/video/BV1rP4y1D7ku),相信结合视频再看本篇题解,更有助于大家对本题的理解**。


## 思路

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![337.打家劫舍III](https://code-thinking-1253855093.file.myqcloud.com/pics/20210223173849619.png)

# 算法公开课

**《代码随想录》算法视频公开课:[动态规划,房间连成树了,偷不偷呢?| LeetCode:337.打家劫舍3](https://www.bilibili.com/video/BV1H24y1Q7sY),相信结合视频再看本篇题解,更有助于大家对本题的理解**。


## 思路

这道题目和 [198.打家劫舍](https://programmercarl.com/0198.打家劫舍.html),[213.打家劫舍II](https://programmercarl.com/0213.打家劫舍II.html)也是如出一辙,只不过这个换成了树。
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* 0 < prices[i] < 50000.
* 0 <= fee < 50000.

# 算法公开课

**《代码随想录》算法视频公开课:[动态规划来决定最佳时机,这次含手续费!| LeetCode:714.买卖股票的最佳时机含手续费](https://www.bilibili.com/video/BV1z44y1Z7UR),相信结合视频再看本篇题解,更有助于大家对本题的理解**。


## 思路

本题贪心解法:[贪心算法:买卖股票的最佳时机含手续费](https://programmercarl.com/0714.买卖股票的最佳时机含手续费.html)
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}
```

```java
class Solution {
public int largestSumAfterKNegations(int[] A, int K) {
if (A.length == 1) return k % 2 == 0 ? A[0] : -A[0];
Arrays.sort(A);
int sum = 0;
int idx = 0;
for (int i = 0; i < K; i++) {
if (i < A.length - 1 && A[idx] < 0) {
A[idx] = -A[idx];
if (A[idx] >= Math.abs(A[idx + 1])) idx++;
continue;
}
A[idx] = -A[idx];
}

for (int i = 0; i < A.length; i++) {
sum += A[i];
}
return sum;
}
}
```

### Python
```python
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<img src='https://code-thinking.cdn.bcebos.com/pics/动态规划-总结大纲1.jpg' width=600> </img>

## 算法公开课

**《代码随想录》算法视频公开课:[动态规划理论基础](https://www.bilibili.com/video/BV13Q4y197Wg),相信结合视频再看本篇题解,更有助于大家对本题的理解**。


## 什么是动态规划

动态规划,英文:Dynamic Programming,简称DP,如果某一问题有很多重叠子问题,使用动态规划是最有效的。
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AltStyle によって変換されたページ (->オリジナル) /