Best Time to Buy and Sell Stock - LeetCodeblog
解法一:遍歷一遍,隨時記錄當前數字以前的最小的數字。將當前數字與當前最小數字相減查看收益。時間複雜度O(n)leetcode
class Solution { public: int maxProfit(vector<int>& prices) { int size = prices.size(); if(size < 1) return 0; int max = 0,min = prices[0]; for(int i = 1;i < size;i++) { if(prices[i] - min > max) max = prices[i] - min; if(prices[i] < min) min = prices[i]; } return max; } };