494. Target Sum（第十一周）

最新推荐文章于 2021-07-29 14:31:32 发布

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探讨如何通过递归算法解决符号组合求和问题，即给定一组非负整数及目标值S，寻找使得整数之和等于S的有效符号组合方式的数量。

Description：

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.

解题思路：

这道题要用到计算所有有可能出现的情况然后再比较，用递归可以做。时间复杂度为O（n^2）。

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