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.

Constraints:

• 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.

class Solution {
    int res = 0;
    public int findTargetSumWays(int[] nums, int S) {
        
 
if(nums.length == 0) return 0;
        dfs(nums, S, 0, 0);
        return res;
    }
    public void dfs(int[] nums, int S, int pos, int cur){
        if(pos == nums.length) {
            if(cur == S) res++;
            return;
        }
        dfs(nums, S, pos + 1, cur + nums[pos]);
        dfs(nums, S, pos 
 
+ 1, cur - nums[pos]);
    }
}

1. dfs, 每一步都可以加減，所以從cur0開始，每次dfs都可以±。

顯然 還有更快的方法

public int findTargetSumWays(int[] nums, int S) {
      
      int sum = 0;
      for(int n: nums){
        sum += n;
      }
      if (S < -sum || S > sum) { return 0;}
        
      int[][] dp = new int[nums.length + 1][ 2 * sum + 1];
      dp[0][0 + sum] = 1; // 0 + sum means 0, 0 means -sum,  check below graph
      for(int i = 1; i <= nums.length; i++){
        for(int j = 0; j < 2 * sum + 1; j++){
          
          if(j + nums[i - 1] < 2  * sum + 1) dp[i][j] += dp[i - 1][j + nums[i - 1]];
          if(j - nums[i - 1] >= 0) dp[i][j] += dp[i - 1][j - nums[i - 1]];
        }
      }
      return dp[nums.length][sum + S];
    }

揹包法，普通0/1揹包是選或不選，這個是選它或者選它的負數，所以dp [ i ][ j ] = dp[i-1][j-nums[i-1]] + dp[i-1][j + nums [ i - 1];

dp[ i ][ j ]意思是：為了達到sum j，使用了前i個元素，所以上面的轉移方程顯然（個鬼）

因為j的範圍是 -sum ---- 0 ----sum，所以我們預設dp[ 0 ][ sum]相當於起點0，所以dp[ 0 ][ 0 ] 就代表-sum，dp 【i】【sum+S】就是我們所要求的。