Target Sum

Target Sum

A.題意

Follow up for “Unique Paths”:

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:
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.

題意大概是給你一個數組的數字然後你可以使用加號或者減號對兩個相鄰的元素求和，最後看有多少種方法能求得目標和。

B.思路

這道題其實就是深度優先搜索吧，每次選擇一個符號最後判斷是不是想要的結果就好了，控制一下dfs的深度保證不超出給定數組長度即可。

C.代碼實現

class Solution {
public:
    int findTargetSumWays(vector<int>& nums, int S) {
        if (nums.empty()) return 0;
        dfs(nums, S, 0, 0);
        return count;
    }
    void dfs(vector<int>& nums, int s, int k, int sum)
    {
        if (k == nums.size())
        {
            if (s == sum) count++;
            return;
        }
        dfs(nums, s, k + 1, sum + nums[k]);
        dfs(nums, s, k + 1, sum - nums[k]);
    }
private :
    int count = 0;
};