LeetCode 494. Target Sum

原题链接在这里：https://leetcode.com/problems/target-sum/description/

题目：

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.

题解：

List some examples. e.g. 1,1,1,1,1.

If all of them are positive, it is 5, all negitative, it is -5. Beyond [-5,5], it can't happen, thus, the ways count is 0.

For this sum question, let dp[i] denotes the sum up to i, the count of ways.

Iterate each num in nums. For num, it could be + or -. say first 1. Then it could -1 or 1. 

The ways to -1 is 1, ways to 1 is 1. It is because dp[0] accumlate to dp[1] and dp[-1].

递推时, 上个可能结果或加或减当前num得到新的结果, 不同ways的数目在新结果下累计.  只有对应count大于0时才可能是上个可能结果, because it would not be out of index.

起始值dp[0 + sum] = 1. 可能结果为0的不同ways数目是1. sum is offset. dp[0] means sum up to -sum.

Time Complexity: O(sum * n). sum是nums所有num的和. n = nums.length

Space: O(sum).

AC Java:

class Solution {
    public int findTargetSumWays(int[] nums, int S) {
        if(nums == null || nums.length == 0){
            return 0;
        }
        
        int sum = 0;
        for(int num : nums){
            sum += num;
        }
        
        if(sum < S || -sum > S){
            return 0;
        }
        
        int [] dp = new int[2*sum+1];
        //sum相当于 offset
        dp[0+sum] = 1; 
        for(int num : nums){
            int [] next = new int[2*sum+1];
            for(int k = 0; k<2*sum+1; k++){
                if(dp[k] > 0){
                    next[k+num] += dp[k];
                    next[k-num] += dp[k];
                }
            }
            dp = next;
        }
        return dp[sum + S];
    }
}

Method 2:

nums中一部分用的+号 相当于positive, 另一部分用的 - 号相当于negative. 分成两组.

sum(p) - sum(n) = target.

sum(p) + sum(n) + sum(p)-sum(n) = target + sum(p) + sum(n)

2*sum(p) = target + sum(nums)

相当于在nums中找有多少种subarray, subarray自身的和是(target + sum(nums))/2的问题.

subSum求解这个转化问题.

存储到当前数字得到所有结果ways数目. 为什么循环中i要从大到小呢. 这其实是dp的space compression.

update新的dp时会用到上一轮前面的值，所以要从后往前更新. 这样更新时保证用到的都是上轮的值.

Time Complexity: O(sum*nums.length). sum是nums所有num的和.

Space: O(sum).

AC Java:

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

类似Partition Equal Subset Sum.