Lintcode 买卖股票的最佳时机

买卖股票的最佳时机

思路： 最大利润 为当前值- 前面最小值
1 lowest =prices[0],记为最小值，循环一次，记min（lowest，prices[i]）为最小值；
2 profit= max（profit，prices[i] - lowest）;profit 初值为0；

class Solution {public:    /**     * @param prices: Given an integer array     * @return: Maximum profit     */    int maxProfit(vector<int> &prices) {        // write your code here        int m = prices.size();        int profit = 0;        if (m < 2) {            return profit;        }        int lowest = prices[0];        for (int i = 0; i < m; i++) {            int current = prices[i];            profit = max(profit, current - lowest);            lowest = min(current, lowest);        }        return profit;    }};

买卖股票的最佳时机||

class Solution {public:    /**     * @param prices: Given an integer array     * @return: Maximum profit     */    int maxProfit(vector<int> &prices) {        // write your code here        int m = prices.size();        int profit = 0;        if (m < 2) {            return 0;        }        for (int i = 0; i < m-1; i++) {            int j = prices[i+1] - prices[i];            if (j > 0) {                profit += j;            }        }        return profit;    }};

买卖股票的最佳时机3

分析：动态规划法。以第i天为分界线，计算第i天之前进行一次交易的最大收益preProfit[i]，和第i天之后进行一次交易的最大收益postProfit[i]，最后遍历一遍，max{preProfit[i] + postProfit[i]} (0≤i≤n-1)就是最大收益。第i天之前和第i天之后进行一次的最大收益求法同Best Time to Buy and Sell Stock I

没有通过？？？？？

vs上面是编译通过的，结果也正确。

#include<iostream>#include<vector>#include<string>#include<algorithm>using namespace std;class Solution {public:    /**    * @param prices: Given an integer array    * @return: Maximum profit    */    int maxProfit(vector<int> &prices) {        if (prices.size() < 2) return 0;        int n = prices.size();        vector<int> preProfit(n,0);        vector<int> postProfit(n,0);        int curMin = prices[0];        for (int i = 1; i < n; i++) {            curMin = min(curMin, prices[i]);            preProfit[i] = max(preProfit[i - 1], prices[i] - curMin);        }        int curMax = prices[n - 1];        for (int i = n - 2; i >= 0; i--) {            curMax = max(curMax, prices[i]);            postProfit[i] = max(postProfit[i + 1], curMax - prices[i]);        }        int maxProfit = 0;        for (int i = 0; i < n; i++) {            maxProfit =max(maxProfit, preProfit[i] + postProfit[i]);        }        return  maxProfit;    }};int main(){    Solution test;//我们取一个名字为test的对象    cout << "请输入一串整数数组" << endl;    int num;    vector<int>b;    while (cin >> num)    {        b.push_back(num);        if (cin.get() == '\n')   //如果检测到用户回车，则结束输入            break;    }    test.maxProfit(b);    cout << test.maxProfit(b) << endl;    system("pause");    return 0;}