LeetCode

You are given a list of non-negative integers, a1, a2, ..., an, and a target, S. Now you have 2 symbols + and -. For each integer, you should choose one from + and - as its new symbol.

Find out how many ways to assign symbols to make sum of integers equal to target S.

Example 1:

Input: nums is [1, 1, 1, 1, 1], S is 3. Output: 5Explanation: -1+1+1+1+1 = 3+1-1+1+1+1 = 3+1+1-1+1+1 = 3+1+1+1-1+1 = 3+1+1+1+1-1 = 3There are 5 ways to assign symbols to make the sum of nums be target 3.

Note:

1. The length of the given array is positive and will not exceed 20.
2. The sum of elements in the given array will not exceed 1000.
3. Your output answer is guaranteed to be fitted in a 32-bit integer.

一碰到dp，就很容易懵。写写正解思路吧，参考  http://blog.csdn.net/mine_song/article/details/70216562

【问题分析】

1、该问题求解数组中数字只和等于目标值的方案个数，每个数字的符号可以为正或负(减整数等于加负数)。

2、该问题和矩阵链乘很相似，是典型的动态规划问题

3、举例说明: nums = {1,2,3,4,5}, target=3, 一种可行的方案是+1-2+3-4+5 = 3

该方案中数组元素可以分为两组，一组是数字符号为正(P={1,3,5})，另一组数字符号为负(N={2,4})

因此:    sum(1,3,5) - sum(2,4) = target

sum(1,3,5) - sum(2,4) + sum(1,3,5) + sum(2,4) = target + sum(1,3,5) + sum(2,4)

2sum(1,3,5) = target + sum(1,3,5) + sum(2,4)

2sum(P) = target + sum(nums)

sum(P) = (target + sum(nums)) / 2

由于target和sum(nums)是固定值，因此原始问题转化为求解nums中子集的和等于sum(P)的方案个数问题

4、求解nums中子集和为sum(P)的方案个数(nums中所有元素都是非负)

此处就是一个经典的0/1背包了，详细过程省略

class Solution {public:    int findTargetSumWays(vector<int>& nums, int S) {        int sum = 0;        for (int i = 0; i < nums.size(); ++i)            sum += nums[i];        if (sum < S || (sum + S) % 2 == 1) return 0;        return solve(nums, (sum + S) / 2);    }    int solve(vector<int> nums, int s) {        vector<int> ans(s+1, 0);        ans[0] = 1;        for (int i = 0; i < nums.size(); ++i) {            for (int j = s; j >= nums[i]; --j) {                ans[j] += ans[j-nums[i]];            }        }        return ans[s];    }};