第十二周 动态规划（二)

算法题目 ： Target Sum                   

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算法题目描述： 

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: 5Explanation: -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 = 3There 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.

算法分析：

这道题目是给定一个非负的整数数组，a1,a2,...,an ,以及一个目标值S，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。

用动态规划里面的dp,

sum(P) - sum(N) = target
sum(P) + sum(N) + sum(P) - sum(N) = target + sum(P) + sum(N)
                       2 * sum(P) = target + sum(nums)
所以原问题就转化成了子数组和问题-----找到一个子集P使得 sum(P) = (target + sum(nums)) / 2

算法代码（C++):

int subsetSum(int* nums, int numsSize, int target) {      int ret;      int* dp = (int*)malloc((target+1)*sizeof(int));      memset(dp, 0, (target+1)*sizeof(int));      dp[0] = 1;      for(int i = 0; i < numsSize; ++i)          for(int j = target; j >= nums[i]; --j)              dp[j] += dp[j-nums[i]];      ret = dp[target];      free(dp);      return ret;  }    int findTargetSumWays(int* nums, int numsSize, int S) {      int sum = 0, target, ret;      for(int i = 0; i < numsSize; ++i) sum += nums[i];      if(sum < S || (sum + S)%2) return 0;      target = (sum + S)/2;      ret = subsetSum(nums, numsSize, target);      return ret;  }