494. Target Sum-优快云博客

本文介绍了一种解决LeetCode上494题（目标和）的方法，通过将问题转化为子集和问题，利用动态规划求解数组中元素通过加减运算达到目标值的所有可能组合。

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494. Target Sum

题目描述：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.

题目大意：给定一个数组nums，给定一个target，问数组中的数有多少种方式可以通过+-，得到target。
思路：给定一个数组nums=[1,2,3,4,5]，target=3，一个可能的解是+1-2+3-4+5 = 3.解中的正数子集P = [1, 3, 5] 负数子集 N = [2, 4]，可得sum(P) - sum(N) = target，两边同时加sum(P) + sum(N)，sum(P) + sum(N) + sum(P) - sum(N) = target + sum(P) + sum(N)，简化得 2 * sum(P) = target + sum(nums)，sum(P) = (target + sum(nums)) / 2。

代码

package DP;

/**
* @author OovEver
* 2017/12/25 10:21
*/
public class LeetCode494 {
  public int findTargetSumWays(int[] nums, int S) {
      int sum = 0;
      for (int n : nums) {
          sum += n;
      }
      if (sum < S || (S + sum) % 2 > 0) {
          return 0;
      }
      return subsetSum(nums, (sum + S) / 2);
  }

  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];
  }
}