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Best Time to Buy and Sell Stock III

时间:2015-01-04 17:14:20      阅读:130      评论:0      收藏:0      [点我收藏+]

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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.

Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).



程序包含了Best Time to Buy and Sell Stock I和Best Time to Buy and Sell Stock II


#include<stdio.h>
//Best Time to Buy and Sell Stock I
int maxProfit(int *prices, int n) {
    int i = 0;
    int profit = 0, min = *prices;
    if(n == 0) return 0;
    
    for(i = 1; i < n; i++){
        if(*(prices + i) - min > profit) profit = *(prices + i) - min;
        if(*(prices + i) < min) min = *(prices + i);
    }
    printf("profit:%d\n", profit);
    return profit;
}
//Best Time to Buy and Sell Stock II
int maxProfitMulti(int *prices, int n) {
    int i, j, sum = 0;
    if(n == 0) return 0;
    for(i = 0 ;i < n - 1; i++ ) {
        if(prices[i + 1] > prices[i]) {
            for(j = i; j < n - 1 && prices[j] < prices[j + 1]; j++);
            if(j != i ) {
                sum += prices[j] - prices[i];
                //printf("%d\n", sum);
            }
            i = j;
        }
    }
    return sum;

}
//Best Time to Buy and Sell Stock III
int maxProfitTow(int *prices, int n) {
    int i, max = 0, profit = 0;
    if(n == 0) return 0;
    for(i = 1 ;i < n; i++ ) {
        if(pricess[i] <= pricess[i - 1]) continue;
        profit = maxProfit(prices, i + 1) + maxProfit(prices + i + 1, n - i - 1);
        if(profit > max) max = profit;
    }
    return max;
}

void main() {
    int prices[] = {1,2,4,2,5,7,2,4,9,0};
    int n = 10;
    printf("%d\n", maxProfitTow(prices, n));
}


Best Time to Buy and Sell Stock III

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原文地址:http://blog.csdn.net/uj_mosquito/article/details/42393491

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