leetcode 121.买卖股票的最佳时机
class Solution {
public:
int maxProfit(vector<int>& prices) {
int ans = 0;
vector<int> St;
prices.emplace_back(-1); //为了结果的必然出现
for (int i = 0; i < prices.size(); ++ i){
while (!St.empty() && St.back() > prices[i]){//维护这个递增栈
ans = std::max(ans, St.back() - St.front());
St.pop_back();
}
St.emplace_back(prices[i]);
}
return ans;
}
};
标签:emplace,int,位置,back,St,查找,差值,ans,prices
From: https://www.cnblogs.com/skiesclear-639/p/17631112.html