LeetCode -- Target Sum

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探讨了给定一组非负整数并使用加减符号使整数之和等于目标值的方法数。采用递归穷举思想，通过不断尝试为每个整数添加正负号来寻找所有可能的组合。

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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: 5
Explanation: 

-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 = 3

There 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.

解读：

给与一组非负整数，给每个整数前加上+或-，加完符号之后的整数相加起来，计算结果为S的组合数。

思路：递归+穷举思想。利用递归实现穷举。确定好前n个数字的符号再确定下一个数字的符号，每确定一个符号更新一次和temp

当数字的数量为给定数组数量的时候判断和是否为目标和，若是result++；否则更改当下数字符号；更新完两次返回上一次数字进行更新。

代码：

class Solution {
public:
    
    void dfs(int S, vector<int>& nums, int& temp, int& result, int depth){
        if(temp == S&&depth == nums.size()) {
            result++;
            return;
        }
        if(depth < nums.size()){
            
                temp += nums[depth];
                dfs(S, nums, temp, result, depth+1);
                temp -= 2*nums[depth];
                dfs(S, nums, temp, result, depth+1);
                temp += nums[depth];
        }
    }
    
    int findTargetSumWays(vector<int>& nums, int S) {
        int temp = 0;
        int result = 0;
        int depth = 0;
        dfs(S, nums, temp, result, depth);
        
        return result;
    }
    
};