leetcode 121. Best Time to Buy and Sell Stock

最新推荐文章于 2025-04-18 22:24:21 发布

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最新推荐文章于 2025-04-18 22:24:21 发布

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本文介绍了一种寻找股票买卖最佳时机以获得最大利润的算法。通过两种不同的方法实现：一种是记录每天卖出的最大利润；另一种则借鉴了Kadane算法的思想，计算连续子数组的最大利润。适用于单一交易。

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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.

使用一个数组保存每一天卖出时能产生的最大利润，当价格小于最小值时覆盖最小值，最后遍历该数组得到最大利润

public class Solution {
    public int maxProfit(int[] prices) {
        int len = prices.length;
        if(len==0 || len==1) return 0;
        int ans=0;
        int i=0, j=1; 
        int[] max = new int[len];
        for(; j<len;j++){
            if(prices[j]>prices[i]){
                max[j] = prices[j] - prices[i];
            }
            if(prices[j]<prices[i]){
                i = j;
            }
        }
        for(int k=1; k<len; k++){
            if(max[k]>ans) ans = max[k];
        }
        return ans;
    }
}

好像也不需要那么复杂

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