494. Target Sum_you are given a non-empty list of positive integer-优快云博客

本文链接：https://blog.youkuaiyun.com/guo_x_l/article/details/72870420

本文探讨了一种特定的问题——如何通过给定的一组非负整数，并使用加号和减号来达到特定的目标和。文章提供了详细的解决方案及示例代码，采用动态规划的方法解决了这一问题。

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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:

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.

题解：

这题很容易就能用暴力（深度搜索）的解法。但不是最好的解法，可以用动态规划来解，DP 加入“+”和“-”之后原数组的元素可以看成被分成了两个子集合，一个正的子集合P，另一个负的子集合N。使得sum(P)+sum(N) = S，且P+N = nums。不断地简化，最终只要使得sum(P) = (S + sum(nums)) / 2，即可。

具体代码如下：

class Solution {
public:
    int findTargetSumWays(vector<int>& nums, int S) {
        int sum = accumulate(nums.begin(), nums.end(), 0);
        if (sum < S || ((S + sum) & 0x1) == 1) {
            return 0;
        }
        int target = (S + sum) >> 1;
        vector<int> dp(target + 1, 0);
        dp[0] = 1;
        for (auto num : nums) {
            for (int i = target; i >= num; --i) {
                dp[i] += dp[i - num];
            }
        }
        return dp[target];
      }
};

end!