LeetCode 122. 买卖股票的最佳时机 II

就是不断寻找最长的非递增序列

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if(prices.empty()||prices.size()<2){
            return 0;
        }
        int i=0;
        int res = 0;
        int j=0 ;
        int size = prices.size();
        while(i<size){
            j = i+1;
            int pre = prices[i];
            while(j<size&&prices[j]>=pre){
                pre = prices[j];
                j++;
            }
            if(i+1!=j){
                res += prices[j-1] - prices[i];
            }
            i = j;
        }
        return res;
    }
};

原文地址：http://www.cnblogs.com/poteitoutou/p/16877868.html