494. Target Sum
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:
- 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.
题目含义：给了一个非负数的矩阵，里面的元素是[a0, a1, a2……]和一个目标值S，现在要在这些元素之间加入+或者-，使得里面的元素之间运算结果能够等于目标值S，求一共有多少种加入+或者-的方法
1 class Solution { 2 3 4 int subsetSum(int[] nums, int s) { 5 // dp[i]表示子集合元素之和等于当前目标值的方案个数, 当前目标值等于9减去当前元素值 6 // 当前元素等于1时， dp[9] = dp[9] + dp[9-1] 7 // dp[8] = dp[8] + dp[8-1] 8 // ... 9 // dp[1] = dp[1] + dp[1-1] 10 // 当前元素等于2时，dp[9] = dp[9] + dp[9-2] 11 // dp[8] = dp[8] + dp[8-2] 12 // ... 13 // dp[2] = dp[2] + dp[2-2] 14 // 当前元素等于3时，dp[9] = dp[9] + dp[9-3] 15 // dp[8] = dp[8] + dp[8-3] 16 // ... 17 // dp[3] = dp[3] + dp[3-3] 18 int[] dp = new int[s + 1]; 19 dp[0] = 1; 20 for (int n : nums) 21 for (int i = s; i >= n; i--) 22 dp[i] += dp[i - n]; 23 return dp[s]; 24 } 25 26 27 // 该方案中数组元素可以分为两组，一组是数字符号为正(P={1,3,5})，另一组数字符号为负(N={2,4}) 28 // 因此: sum(1,3,5) - sum(2,4) = target 29 // sum(1,3,5) - sum(2,4) + sum(1,3,5) + sum(2,4) = target + sum(1,3,5) + sum(2,4) 30 // 2sum(1,3,5) = target + sum(1,3,5) + sum(2,4) 31 // 2sum(P) = target + sum(nums) 32 // sum(P) = (target + sum(nums)) / 2 33 // 由于target和sum(nums)是固定值，因此原始问题转化为求解nums中子集的和等于sum(P)的方案个数问题 34 public int findTargetSumWays(int[] nums, int S) { 35 int sum = 0; 36 for (int n : nums) 37 sum += n; 38 return sum < S || (S + sum) % 2 > 0 ? 0 : subsetSum(nums, (S + sum) / 2); 39 } 40 }
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