题目:

代码(首刷看解析 2024年2月26日)
思路:根据题意,设两个背包,packageA存放前面是"+"的数字之和,packageB存放前面是"-"的数字之和
则sum = packageA + packageB;
target = packageA - packageB;
则根据以上算式可得packageA = (sum + target)/2;
因为sum和target根据题目都已知,则问题变成求可以放哪些数使得和 == packageA--即动态规划01背包问题
            
            
              cpp
              
              
            
          
          class Solution {
public:
    int findTargetSumWays(vector<int>& nums, int target) {
        int sum = 0;
        for (int& num :nums) {
            sum += num;
        }
        if (sum < abs(target)) return 0; 
        int tmp = sum + target;
        if (tmp % 2 == 1) return 0;
        int x = tmp / 2;
        vector<int> dp(x + 1, 0);
        dp[0] = 1;
        for (int i = 0; i < nums.size(); ++i) {
            for (int j = x; j >= nums[i]; --j) {
                dp[j] += dp[j - nums[i]];
            }
        }
        return dp[x];
    }
};