• LeetCode 494. Target Sum


    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.

    分析

    题目的意思等价于把题目中给定的集合中挑选元素分成两个集合,set1,set2,然后用set1的和减去set2的和等于S。
    set1-set2=S;
    set1-(sum-set1)=S;
    set1=(S+sum)/2;
    所以题目转化成在集合中挑选元素使其和为(S+sum)/2.

    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:Sumset(nums,(S+sum)/2);
        }
        int Sumset(vector<int>& nums,int s){
            int dp[s+1]={0};
            dp[0]=1;
            for(auto num:nums)
                for(int i=s;i>=num;i--)
                    dp[i]+=dp[i-num];
            return dp[s];
        }
    };
    
    

    这道题用dfs也可以,尝试为每个数前面添加 + 或 -。

    class Solution {
    public:
        int res=0;
        void dfs(int cnt,int sum,vector<int>& nums,int S){
            if(cnt==nums.size()){
                if(sum==S)
                 res++;  
                 return ;
            }      
            dfs(cnt+1,sum+nums[cnt],nums,S);
            dfs(cnt+1,sum-nums[cnt],nums,S);
        }
        int findTargetSumWays(vector<int>& nums, int S) {
            dfs(0,0,nums,S);
            return res;
        };
    };
    
    
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  • 原文地址:https://www.cnblogs.com/A-Little-Nut/p/10061285.html
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