Leetcode每日一题：121.best-time-to-buy-and-sell-stock(买股票的最佳时机)

暴力法：O(n²)复杂度，即双循环，把所有的对数都比一遍，但肯定超时通过不了；

双指针递归法：左指针和右指针分别从左端和右端开始，如果prices[l]<min 那么i++ 且 min=prices[i] ；右端同理，然后做差进行比较得到result，随后递归；对 ( l+1,r ) 和 ( l , r-1 )重复计算；但仍然超时；

    void Profit(vector<int> &prices, int l, int r, int &result){if(l>=r) return;int min = prices[l], max = prices[r];if (l < r){l++;while (l < r && prices[l] < min){min = prices[l];l++;}r--;while (l < r && prices[r] > max){max = prices[r];r--;}}l--;r++;if (max > min && max - min > result)result = max - min;Profit(prices, l, r - 1, result);Profit(prices, l + 1, r, result);}int maxProfit(vector<int> &prices){int l = 0, r = prices.size() - 1;int result = 0;Profit(prices, l, r, result);return result;}

动态规划：得到评论区的一条评论，瞬间领悟；动态规划前i天的最大收益 = max{前i-1天的最大收益，第i天的价格-前i-1天中的最小价格}，这样的话O(n)的复杂度即可完成；

int maxProfit(vector<int> &prices)
{int result = 0; //前i-1天的最大收益int len = prices.size();if (len < 2)return 0;int min = prices[0];for (int i = 1; i < len; i++){if (prices[i] < min){min = prices[i];continue;}if (prices[i] - min > result)result = prices[i] - min;}return result;
}

func maxProfit(prices []int) int {len := len(prices)if len == 0 {return 0}res := 0start, end := 0, 1for ; end < len; end++ {if prices[end] < prices[end-1] {res += prices[end-1] - prices[start]start = end}}//注意别漏判了end==len的情况，此时start到end-1这一段都是递增的，应加入resif end == len {res += prices[end-1] - prices[start]}return res
}

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