[LeetCode] (medium) 494. Target Sum

https://leetcode.com/problems/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:

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.

---

DFS，直观但慢

class Solution {
public:
    vector<int> remain;
    int result = 0;
    
    int findTargetSumWays(vector<int>& nums, int S) {
        remain = vector<int>(nums.size(), 0);
        remain.back() = nums.back();
        for(int i = nums.size()-2; i >= 0; --i){
            remain[i] = remain[i+1]+nums[i];
        }
        
        dfs(0, 0, S, nums);
        return result;
    }
    
    void dfs(int pos, int inter, int S, vector<int>& nums){
        if(pos == nums.size()){
            if(inter == S) ++result;
            return;
        }else{
            if(S > inter+remain[pos] || S < inter-remain[pos]) return;
            dfs(pos+1, inter+nums[pos], S, nums);
            dfs(pos+1, inter-nums[pos], S, nums);
        }
    }
};

 

化为子集合问题，使用dp求解，快

https://leetcode.com/problems/target-sum/discuss/97334/Java-(15-ms)-C%2B%2B-(3-ms)-O(ns)-iterative-DP-solution-using-subset-sum-with-explanation

将正数和负数看成两个数集

class Solution {
public:
    int findTargetSumWays(vector<int>& nums, int S) {
        int sum = 0;
        for(int n : nums) sum += n;
        
        if(sum < S || (S+sum)%2 != 0) return 0;
        
        int target = (sum+S)/2;
        vector<int> dp(target+1, 0);
        dp[0] = 1;
        
        for(int n : nums){
            for(int i = target; i >= n; --i){
                dp[i] += dp[i-n];
            }
        }
        
        return dp[target];
    }
};