494. Target Sum

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题目：

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: #include<iostream>#include<vector>#include<numeric>using namespace std;    class Solution {public:    int findTargetSumWays(vector<int>& nums, int s) {        int sum = accumulate(nums.begin(), nums.end(), 0);        return sum < s || (s + sum) %2 > 0 ? 0 : subsetSum(nums, (s + sum) >> 1);     }       int subsetSum(vector<int>& nums, int s) {        int dp[s + 1] = { 0 };        dp[0] = 1;        for (int j = 0; j < nums.size(); j++ )            for (int i = s; i >= nums[j]; i--)                dp[i] += dp[i - nums[j]];        return dp[s];    }};int main() {int arr[] = {1,1,1,1,1};vector<int> v(arr,arr+5);Solution s; cout << s.findTargetSumWays(v,3)<<endl;}

-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:

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

翻译：

你有一个非负整数列表，A1，A2，A，A和目标，S.现在你有2个符号+和—。对于每个整数，你应该选择一个从+和-作为它的新符号。
找出分配符号使整数等于目标S的方法有多少种
例1：
输入：公司是[ 1，1，1，1，1 ]，S为3。
输出：5
解释：
- 1 + 1 + 1 + 1 + 1 = 3
+1 + 1 + 1 + 1 = 3
+ 1 + 1-1 + 1 + 1 = 3
+ 1 + 1 +1 +1 + 1 = 3
+ 1 + 1 + 1 + 1-1 = 3
有5种方法来指定符号让你和是目标3。
注：
给定数组的长度为正，不会超过20。
给定数组中元素的总和不超过1000。
您的输出答案是保证在一个32位整数。

解题思路：

1.可以将该序列arr分为正整数和负整数部分集合，分别用ps，和ng表示，那么target = ps - ng，sum表示该序列绝对值的和

2.target +sum = ps - ng +ps + ng = 2 ps；

3. 求该序列中有多少项满足 ps = （target +sum）/2

解题代码：

#include<iostream>#include<vector>#include<numeric>using namespace std;    class Solution {public:    int findTargetSumWays(vector<int>& nums, int s) {        int sum = accumulate(nums.begin(), nums.end(), 0);        return sum < s || (s + sum) %2 > 0 ? 0 : subsetSum(nums, (s + sum) >> 1);     }       int subsetSum(vector<int>& nums, int s) {        int dp[s + 1] = { 0 };        dp[0] = 1;        for (int j = 0; j < nums.size(); j++ )            for (int i = s; i >= nums[j]; i--)                dp[i] += dp[i - nums[j]];        return dp[s];    }};int main() {int arr[] = {1,1,1,1,1};vector<int> v(arr,arr+5);Solution s; cout << s.findTargetSumWays(v,3)<<endl;}

解题状态：

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