Leetcode/G家F家 -- 494. Target Sum (Backtracking)

494. Target Sum (Backtracking)
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相关题： http://rainykat.blogspot.com/2017/01/leetcodegf-282-expression-add.html

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.

思路： use dfs to find travel the array, each number have 2 opitons: '+' and '-'
Complexity: O(2^n)

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

 Optimization: If the sum of all elements left is smaller than absolute value of target, there will be no answer following the current path. Thus we can return.

public class Solution {
    public int findTargetSumWays(int[] nums, int S) {
        if(nums == null)return 0;
        int n = nums.length;
        int[] sums = new int[n];
        sums[n-1] = nums[n-1];
        for(int i = n-2; i >= 0; i--){
            sums[i] = nums[i] + sums[i+1];
        }
        return dfs(nums, sums, S, 0, 0);
    }
    public int dfs(int[] nums, int[] sums, int S, int index, int sum){
        int res = 0;
        if(index == nums.length){
            if(sum == S) res++;
            return res;
        }
        if(sums[index] < Math.abs(S - sum))return 0;
        res += dfs(nums, sums, S, index + 1, sum + nums[index]);
        res += dfs(nums, sums, S, index + 1, sum - nums[index]);
        return res;
    }
}