Leetcode 121. Best Time to Buy and Sell Stock

121. Best Time to Buy and Sell Stock

• Total Accepted: 115636
• Total Submissions: 313848
• Difficulty: Easy

Say you have an array for which the ith 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.

Example 1:

Input: [7, 1, 5, 3, 6, 4]
Output: 5

max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)

Example 2:

Input: [7, 6, 4, 3, 1]
Output: 0

In this case, no transaction is done, i.e. max profit = 0.

思路：DP。

方法一：对于第i天，找到[0,i-1]天买进的最低价格，从而得到第i天出售的最高价格。然后遍历i，得到最高的出售价格。

方法二：最大子序列和问题。

代码：

方法一：

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n=prices.size(),i,maxprofit=0,minprice=INT_MAX;
        for(i=0;i<n;i++){
            if(maxprofit<prices[i]-minprice){
                maxprofit=prices[i]-minprice;
            }
            if(minprice>prices[i]){
                minprice=prices[i];
            }
        }
        return maxprofit;
    }
};

方法二：

参考：

https://discuss.leetcode.com/topic/19853/kadane-s-algorithm-since-no-one-has-mentioned-about-this-so-far-in-case-if-interviewer-twists-the-input

https://en.wikipedia.org/wiki/Maximum_subarray_problem

原文：

The logic to solve this problem is same as "max subarray problem" using Kadane's Algorithm. Since no body has mentioned this so far, I thought it's a good thing for everybody to know.

All the straight forward solution should work, but if the interviewer twists the question slightly by giving the difference array of prices, Ex: for {1, 7, 4, 11}, if he gives {0, 6, -3, 7}, you might end up being confused.

Here, the logic is to calculate the difference (maxCur += prices[i] - prices[i-1]) of the original array, and find a contiguous subarray giving maximum profit. If the difference falls below 0, reset it to zero.

    public int maxProfit(int[] prices) {
        int maxCur = 0, maxSoFar = 0;
        for(int i = 1; i < prices.length; i++) {
            maxCur = Math.max(0, maxCur += prices[i] - prices[i-1]);
            maxSoFar = Math.max(maxCur, maxSoFar);
        }
        return maxSoFar;
    }
*maxCur = current maximum value

*maxSoFar = maximum value found so far

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int i,n=prices.size(),curprofit=0,maxprofit=0;
        for(i=1;i<n;i++){
            curprofit=max(0,curprofit+prices[i]-prices[i-1]);
            maxprofit=max(maxprofit,curprofit);
        }
        return maxprofit;
    }
};