leetcode -- 494. Target Sum【数学转化 + 动态规划】

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.

题意

给定一个非负的整数数组，a1,a2,...,an ,以及一个目标值S。现在你又两个符号 + 和 - 。对于每一个整数，你需要从 + 或者 - 中选择一个符号。

找出有多少种符号分配方式使得整数的和等于目标S。

分析及解答

来自：Java (15 ms) C++ (3 ms) O(ns) iterative DP solution using subset sum with explanation

• 【 与数学的关系 】作者首先用数学公式将问题表述出来，根据正，负划分成了两个集合。然后考虑 target 与 两个集合之间构成的约束的关系。然后作者使用数学技巧，将其转化为2 * sum(P) = target + sum(nums)，这说明原问题变成了，只和正数的和，目标值target,所有数字的和sums相关的问题。
• 【 分析数学公式 】在数学公式中，target是常量。 进一步转化得到sum(P) = (target + sum(nums)) / 2，sum(nums)也是容易计算的。这个公式就避免了对于sum(P)的计算。
• 
  

The recursive solution is very slow, because its runtime is exponential

The original problem statement is equivalent to:
Find a subset of nums that need to be positive, and the rest of them negative, such that the sum is equal totarget

Let P be the positive subset and N be the negative subset
For example:
Given nums = [1, 2, 3, 4, 5] and target = 3 then one possible solution is+1-2+3-4+5 = 3
Here positive subset is P = [1, 3, 5] and negative subset isN = [2, 4]

Then let's see how this can be converted to a subset sum problem:

                  sum(P) - sum(N) = target
sum(P) + sum(N) + sum(P) - sum(N) = target + sum(P) + sum(N)
                       2 * sum(P) = target + sum(nums)

So the original problem has been converted to a subset sum problem as follows:
Find a subset P of nums such that sum(P) = (target + sum(nums)) / 2

Note that the above formula has proved thattarget + sum(nums) must be even
We can use that fact to quickly identify inputs that do not have a solution (Thanks to@BrunoDeNadaiSarnaglia for the suggestion)
For detailed explanation on how to solve subset sum problem, you may refer to Partition Equal Subset Sum

Here is Java solution (15 ms)

    public int findTargetSumWays(int[] nums, int s) {
        int sum = 0;
        for (int n : nums)
            sum += n;
        return sum < s || (s + sum) % 2 > 0 ? 0 : subsetSum(nums, (s + sum) >>> 1); 
    }   

    public int subsetSum(int[] nums, int s) {
        int[] dp = new int[s + 1]; 
        dp[0] = 1;
        for (int n : nums)
            for (int i = s; i >= n; i--)
                dp[i] += dp[i - n]; 
        return dp[s];
    }