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.
用一个min表示当前数据之前的最小值,最大利益,就是当前价格跟前面最小值只差的最大值。动态规划。可以使用一个数组表示每个位置的最大利益(与前面最小值的差),但是由于只要求出这些最大利益的最大值就行,所以不用数组,用一个变量就行。
class Solution { public int maxProfit(int[] prices) { if(prices==null||prices.length<2) return 0; int maxProfit=0; int min=prices[0]; for(int i=1;i<prices.length;i++){ if(prices[i]-min>maxProfit) maxProfit=prices[i]-min; if(min>prices[i]) min=prices[i]; } return maxProfit; } }