买卖股票的最佳时机||

Posted 2020-09-29

代码（C++）:

class Solution {
/**
* @param prices: Given an integer array
* @return: Maximum profit
*/
public int maxProfit(int[] prices) {
// write your code here
if (0 == prices.length || null == prices) return 0;
int minPrice = prices[0];
int sumProfit = 0;
for (int i = 1; i < prices.length; i++) {
if (minPrice < prices[i]) {
sumProfit += prices[i] - minPrice;
minPrice = prices[i];
} else {
minPrice = prices[i];
}
}
return sumProfit;
}

};