LeetCode- Best Time to Buy and Sell Stock

LeetCode- Best Time to Buy and Sell Stock

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题目描述

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.

只能买卖至多一次，求最大利润。

算法理解

1. 遍历数组，记录最小值和当前最大值。

   • 如果当前价格小于最小值，更新，
     当前价格减最小值，如果大于之前的最大值，更新最大值。

2. ob大神视角(Kadane’s algorithm(Kadane算法))

[1, 7, 4, 11] =》 [0,6,-3, 7];(difference array of prices)

diff[0] = 0;

等价于一个经典问题。
Maximum of subarray(最大子序列和)

Maximum of subarray(最大子序列和)
找到一个一维数组内的连续的子数组的最大数字的总和。

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Kadane’s algorithm
详情看我的博客页。

*maxCur = current maximum value

*maxSoFar = maximum value found so far

（有时间回来补充）

代码块

代码一

int maxProfit(vector<int> &prices) {
    int maxPro = 0;
    int minPrice = prices[0];
    for(int i = 1; i < prices.size(); i++){
        minPrice = min(minPrice, prices[i]);
        maxPro = max(maxPro, prices[i] - minPrice);
    }
    return maxPro;
}

代码二

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;
    }