# 122.买卖股票的最佳时机II

## 问题描述
[122.买卖股票的最佳时机II](https://leetcode.cn/problems/best-time-to-buy-and-sell-stock-ii/)

## 解题思路
本题的关键是要找`dp`的递推关系，分两种情况讨论:
- `prices[i - 1]`不会被选择，那么`dp[i] = dp[i - 1]`，其实也说明，`prices[i - 1] < prices[i - 2]`；
- `prices[i - 1]`被选择，那么`prices[i - 1] >= prices[i - 2]`，`dp[i] = dp[i - 1] + prices[i - 1] - prices[i - 2]`。

## 代码
```cpp
class Solution {
  public:
    int maxProfit(vector<int> &prices) {
        if (prices.size() == 1)
            return 0;
        vector<int> dp(prices.size() + 1, 0);
        for (int i = 2; i <= prices.size(); i++) {
            if (prices[i - 1] >= prices[i - 2])
                dp[i] = dp[i - 1] + prices[i - 1] - prices[i - 2];
            else
                dp[i] = dp[i - 1];
        }
        return dp[prices.size()];
    }
};
```