leetcode-121

Best Time to Buy and Sell Stock

题目

Say you have an array for which the i^th element is the price of a given stock on dayi.

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: 5max. 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: 0In this case, no transaction is done, i.e. max profit = 0.

解题思路

给出一个股票的价格序列，买入一次卖出一次，求能获得最大利润。其实就是求一个数组中，后面的数减去前面的数能得到的最大值。最容易想到的肯定是每次选一个数，遍历后面的数，求出直接的差然后和当前最大值进行比较，这样时间复杂度为O(n*n),但是最后会超时。对题目进行分析，发现可以用动态规划的方法求解。遇到临时最小的，就保存（1），计算后面比它大的差，保留最大（6-1）java代码
public int maxProfit(int[] prices) {        if (prices.length ==0 ) return 0;          int minPrice = prices[0];          int maxGap = 0;          for (int i = 1; i < prices.length; ++i) {              if(prices[i] - minPrice > maxGap) {                  maxGap = prices[i] - minPrice;              }              if (prices[i] < minPrice) {                  minPrice = prices[i];              }          }          return maxGap;     }

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