不相邻的最大子数组和

问题描述

给一个数组，数组元素为不小于零，求和最大的子数组，其中每个元素在原数组中不相邻。

解题思路

刚拿到题目可能隐约觉得是个dp问题，求前i个元素的最大子数组和，但还是有点手足无措，关键是将问题分情况讨论：前i个元素的最大子数组包含第i个元素和不包含第i个元素。

1. 包含第i个元素，则一定不能包含第i-1个元素，包含第i个元素的最大子数组为不包含第i-1个元素的最大子数组和加上第i个元素
2. 不包含第i个元素，则前i个元素的最大子数组和就是前i-1个元素的最大子数组和

设包含第i个元素的最大子数组和为dp[i]，不包含第i个元素的最大子数组和为np[i]，前i个元素的最大子数组和为max(dp[i], np[i])，根据上面的分析，则有如下推导公式：

• dp[i] = np[i-1]+arr[i]
• np[i] = max(np[i-1], dp[i-1])

代码

#include <stdio.h>#include <stdlib.h>/*Problem description:Find the max sub sum in the arr that each element doesn't neighbor each other.Elements in the array are not less than 0.*/int max_sub_sum(int* arr, int len){    if(len<0)    {        printf("error: array length is less than 0!\n");        return -1;    }    int result = 0;    /*    dp[i]: max sub sum that includes the element i    */    int* dp = (int*)malloc(sizeof(*dp)*len);    /*    np[i]: max sub sum that excludes the element i    */    int* np = (int*)malloc(sizeof(*np)*len);    /*    This is a dp problem.    If we use arr[i] to get the result, then we cannot use arr[i-1].    so we get:    dp[i] = np[i-1]+arr[i]    If we don't use arr[i] to get the result, then we the ith max sub sum    equals the (i-1)th max sub sum, for we don't care about whether it includes    arr[i].    so we get:    np[i] = max(dp[i-1], np[i-1])    Time complexity is O(n).    */    dp[0] = arr[0];    np[0] = 0;    int i=0;    for(i=1; i<len; i++){        dp[i] = np[i-1] + arr[i];        np[i] = dp[i-1]>np[i-1] ? dp[i-1] : np[i-1];    }    result = dp[len-1]>np[len-1] ? dp[len-1] : np[len-1];    free(dp);    free(np);    return result;}int main(){    int arr[8] = {1, 7, 4, 0, 9, 4, 8, 8};    int result = max_sub_sum(arr, 8);    printf("result: %d\n", result);    return 0;}

codeblocks源码链接：
https://github.com/lilingyu/max_sub_sum

ref

http://blog.csdn.net/realxie/article/details/8063885