[LeetCode] [LeetCode]Best Time to Buy and Sell Stock

Say you have an array for which the i^th 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.

看了别人的代码才发现我自己写的是在太复杂，很简单的问题写了这么多。。。好吧，知道差距了，任重道远。不过发现我的版本运行时间比较短

//Run Status: Accepted!//Program Runtime: 44 milli secsclass Solution {public:    int maxProfit(vector<int> &prices) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        int size=prices.size();        if(size==0||size==1) return 0;int *newPrices=new int[size];int *maxPrices=new int[size];int *minPrices=new int[size];int n=0;int mx=0;int mn=0;for(int s=0;s<size;s++)//合并相同的点{if(s<size-1&&prices[s]==prices[s+1]) continue;newPrices[n]=prices[s];n++;}if(newPrices[0]<newPrices[1]){minPrices[mn]=newPrices[0];mn++;}for(int i=1;i<n-1;i++){if(newPrices[i]>newPrices[i-1]&&newPrices[i]>newPrices[i+1]){maxPrices[mx]=newPrices[i];mx++;}if(newPrices[i]<newPrices[i-1]&&newPrices[i]<newPrices[i+1]){minPrices[mn]=newPrices[i];mn++;}}if(newPrices[n-1]>newPrices[n-2]){maxPrices[mx]=newPrices[n-1];mx++;}else{minPrices[mn]=newPrices[n-1];mn++;}int MAX=0;for(int i=0;i<mx;i++) {for(int ii=i;ii<mx;ii++)if(maxPrices[ii]-minPrices[i]>MAX) MAX=maxPrices[ii]-minPrices[i];}return MAX;    }};