[LeetCode] Target Sum通过添加+/-获得目标结果

声明：原题目转载自LeetCode，解答部分为原创

 Problem ：

        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.

Solution：

        思路：动态规划类型。通过给数组的每个数值添加+/-符号，可以将数组分为两个部分，分别是正数组pos[ ]和负数组neg[ ]，目标结果为：

        target = sum_of_pos - sum_of_neg

target = sum_of_pos - (sum_of_array - sum_of_pos)

target = 2 * sum_of_pos - sum_of_array

        通过以上推导，我们可以将题意转化成我们比较熟悉的类型，即求解和为某定值的子集的个数。在本题中，对应的目标和指的是sum_of_pos。

求解和为某定值的子集的个数，详见另一篇文章“[LeetCode] Partition Equal Subset Sum划分数组形成两个和相等的子集”。

        代码如下：

#include<iostream>#include<vector>using namespace std;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)return 0;else if((sum + s) % 2 != 0)return 0;elsereturn count(nums, (sum + s) / 2);    }    int count(vector<int>& nums, int target) {    if(nums.size() == 0)    return 0;        vector<int> count_of_sum(target + 1, 0);    count_of_sum[0] = 1;    for(int i = 0 ; i < nums.size() ; i++)    {    int add = nums[i];    for(int j = target; j >= add; j --)    {    count_of_sum[j] += count_of_sum[j - add];}}return count_of_sum[target];    }};int main(){Solution text;vector<int> nums(8,1);cout << text.findTargetSumWays(nums, 4) << endl;return 0;}