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Best Time to Buy and Sell Stock III
Best Time to Buy and Sell Stock III
Say you have an array for which the ith element is the price of a given stock on day i.
Design an algorithm to find the maximum profit. You may complete at most two transactions.
c++版本代码：
class Solution {public: int maxProfit(vector<int> &prices) { int profit = 0, n = prices.size(); if( n==0 ) { return 0; } int l[n], r[n]; memset(l, 0, sizeof(int)*n); memset(r, 0, sizeof(int)*n); int min = prices[0]; for(int i=1; i<n; i++) { l[i] = prices[i] - min > l[i-1] ? prices[i] - min : l[i-1]; min = prices[i] < min ? prices[i] : min; } int max = prices[n-1]; for(int i=n-2; i>=0; i--) { r[i] = max - prices[i] > r[i+1] ? max - prices[i] : r[i+1]; max = prices[i] > max ? prices[i] : max; } for (int i=0; i<n ;i++) { profit = l[i] + r[i] > profit ? l[i] + r[i] : profit; } return profit; }}; Java版本代码：
public class Solution { public int maxProfit(int[] prices) { if(prices == null || prices.length == 0) { return 0; } int n = prices.length; int[] left = new int[n]; int[] right = new int[n]; int min = prices[0]; for(int i=1; i<n; i++) { left[i] = left[i - 1] > prices[i] - min ? left[i - 1] : prices[i] - min; min = min < prices[i] ? min : prices[i]; } int max = prices[n-1]; for(int i=n-2; i>0; i--) { right[i] = right[i + 1] > max - prices[i] ? right[i + 1] : max - prices[i]; max = max > prices[i] ? max : prices[i]; } int profit = 0; for(int i=0; i<n; i++) { profit = profit > left[i] + right[i] ? profit : left[i] + right[i]; } return profit; }}
Best Time to Buy and Sell Stock III
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