leetcode 123. Best Time to Buy and Sell Stock III

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123. Best Time to Buy and Sell Stock III

 

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• Total Accepted: 62311
• Total Submissions: 230292
• Difficulty: Hard

Say you have an array for which the ith element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete at most two transactions.

Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

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 Array Dynamic Programming

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思路：

思想来源于动态规划，如果以arr[i]为第二个投资点，那么，必须找到i-1前面的最大投资收益

so，，，左右各dp一次。。。

 

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int sz = prices.size();
        vector<int> left(sz + 1,0);
        int i;
        int mi = INT_MAX;
        for(i = 0;i < sz;i++){
            if(i != 0){
                left[i] = left[i - 1];
            }
            if(prices[i] < mi){
                mi = prices[i];
            }
            else{
                left[i] = max(left[i],prices[i] - mi);
            }
        }
        
        vector<int> right(sz + 1,0);
        int ma = INT_MIN;
        for(i = sz - 1;i >= 0;i--){
            if(i != sz - 1){
                right[i] = right[i + 1];
            }
            if(prices[i] > ma){
                ma = prices[i];
            }
            else{
                right[i] = max(right[i],ma - prices[i]);
            }
        }
        //for(i = 0;i < sz;i++){
        //    printf(" i = %d left = %d right = %d
",i,left[i],right[i]);
        //}
        
        int ma_profile = left[sz - 1];
        for(i = 0;i < sz - 1;i++){
            ma_profile = max(ma_profile,left[i] + right[i + 1]);
        }
        return ma_profile;
    }
};