7---LeetCode【tag: Array】【Stock I】|C语言|总结

Say you have an array for which the ith element is the price of a given stock on day i.

If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Example 1:

Input: [7, 1, 5, 3, 6, 4]Output: 5max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)

Example 2:

Input: [7, 6, 4, 3, 1]Output: 0In this case, no transaction is done, i.e. max profit = 0.

问题分析：

看半天没明白这题目什么意思，我还以为是自己阅读理解差，看了讨论才晓得

Please explain the problem more clearly!!!

翻译下就是：

1. 只能先买后卖，并且都只能进行一次；

2. 返回最大的利润

代码一：

int maxProfit(int* prices, int pricesSize) {        int buy = prices[0];    int profit = 0;    for (int i = 0; i < pricesSize; i++) {        if (prices[i] < buy)            buy = prices[i];        if (profit < prices[i] - buy)            profit = prices[i] - buy;    }    return profit;}

代码二：

int maxProfit(int* prices, int pricesSize) {        int buy = prices[0];    int profit = 0;    for (int i = 0; i < pricesSize; i++) {        buy = (prices[i] < buy)? prices[i]: buy;        profit = (profit < prices[i] - buy)? prices[i] - buy: profit;    }    return profit;}

代码说明：

相当于声明了两个寄存器buy和profit，随着时间推移（即i++），buy和profit不断更新，直到循环结束。

思考：

为什么代码一和代码二的效率不一样，代码二效率更高。