动态规划-494. Target Sum

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: 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.

题目解读：给定一个非负整数数组和一个目标值，你可以给这些数字赋予正负号。问有多少种赋值方式可以使这些数字的和等于目标值

DFS Soulutionclass Solution {    public int findTargetSumWays(int[] nums, int S) {        return helper(nums,S,0);    }    public int helper(int[] nums,int target,int start){        if(start == nums.length){            //结束条件            if(target==0) return 1;            else return 0;        }        int count = 0;        count += helper(nums,target-nums[start],start+1);//+号情况        count += helper(nums,target+nums[start],start+1);//-号情况        return count;    }}

//DP Soulution/*Let P be the positive subset and N be the negative subset                  sum(P) - sum(N) = targetsum(P) + sum(N) + sum(P) - sum(N) = target + sum(P) + sum(N)                       2 * sum(P) = target + sum(nums)该问题可以转化为寻找一个子集合P.使得sum(P) = (target+sum(nums))/2注意target+sum(nums)一定是偶数。这可以作为一个快速判断输入是否有解的依据*/class Solution{     public int findTargetSumWays(int[] nums, int S) {        int sum = 0;        for(int i = 0; i < nums.length; i++){            sum += nums[i];        }        return (S > sum || (sum+S) % 2 > 0)? 0 : subSum(nums,(sum+S)/2);    }    public int subSum(int[] nums, int target){        int[] dp = new int[target+1];        dp[0] = 1;        //元素不可重复        for(int num : nums)            for(int j = target; j >= num; j--)                    dp[j] += dp[j-num];        return dp[target];    }}