LeetCode #494 - Target Sum

最新推荐文章于 2020-12-21 22:16:52 发布

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本文介绍了一个给定非负整数数组的问题，通过为数组中的每个元素分配加号或减号来实现目标和的方法数量。使用深度优先搜索算法解决此问题，并提供了一个C++实现示例。

题目描述：

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.

给定非负数组和目标值s，对数组每个数都可以赋+或-，求有多少张赋值方法使得数组和等于s。采用深度优先搜索，传递对参数有nums，s，当前下标i，当前和sum，分别对nums[i]赋正值和负值，分别递归，直到遍历到数组末尾，判断sum是否等于s。

class Solution {
public:
    int findTargetSumWays(vector<int>& nums, int S) {
        int sum=0;
        int i=0;
        return DFS(nums,S,sum,i);
    }
    
    int DFS(vector<int>& nums, int S, int sum, int i)
    {
        int sum1=sum+nums[i];
        int sum2=sum-nums[i];
        if(i==(nums.size()-1))
        {
            int count=0;
            if(sum1==S) count++;
            if(sum2==S) count++;
            return count;
        }
        return DFS(nums,S,sum1,i+1)+DFS(nums,S,sum2,i+1);
    }
};