解法1 : 记录当天左边最小值,右边最大值
解法2:
最大利润只可能在上升阶段
public class Solution {
public int maxProfit(int prices[]) {
int minprice = Integer.MAX_VALUE;
int maxprofit = 0;
for (int i = 0; i < prices.length; i++) {
if (prices[i] < minprice) {
minprice = prices[i];
} else if (prices[i] - minprice > maxprofit) {
maxprofit = prices[i] - minprice;
}
}
return maxprofit;
}
}
作者:LeetCode-Solution
链接:https://leetcode-cn.com/problems/gu-piao-de-zui-da-li-run-lcof/solution/gu-piao-de-zui-da-li-run-by-leetcode-sol-0l1g/
来源:力扣(LeetCode)
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