494. 目标和( 记忆化 dfs)

 

给你一个整数数组 nums 和一个整数 target 。

向数组中的每个整数前添加 '+' 或 '-' ，然后串联起所有整数，可以构造一个 表达式 ：

• 例如，nums = [2, 1] ，可以在 2 之前添加 '+' ，在 1 之前添加 '-' ，然后串联起来得到表达式 "+2-1" 。

返回可以通过上述方法构造的、运算结果等于 target 的不同 表达式 的数目。

 

示例 1：

输入：nums = [1,1,1,1,1], target = 3
输出：5
解释：一共有 5 种方法让最终目标和为 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
+1 + 1 + 1 + 1 - 1 = 3

示例 2：

输入：nums = [1], target = 1
输出：1

 

提示：

• 1 <= nums.length <= 20
• 0 <= nums[i] <= 1000
• 0 <= sum(nums[i]) <= 1000
• -1000 <= target <= 1000

 

 

 

 

 

 

 

 

 

 

 

class Solution:
    def findTargetSumWays(self, nums: List[int], target: int) -> int:

        memo = {}
        def backtracking(index,cur_cum):
            key = str(index) + '-' + str(cur_cum)
            if key in memo:
                return memo[key]
            if cur_cum == target and index == len(nums):
                memo[key] = 1
                return 1
            if  index >= len(nums):
                memo[key] = 0
                return 0
            
            a = backtracking(index+1,cur_cum+nums[index])
            b = backtracking(index+1,cur_cum-nums[index])
            memo[key] = a+b
            return a+b
        return backtracking(0,0)

 

 

pos + neg = sum

pos-neg = target

neg = (sum-target)/2

 

 

 

 

 

​

作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/target-sum/solution/mu-biao-he-by-leetcode-solution-o0cp/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

 

class Solution {
public:
    int findTargetSumWays(vector<int>& nums, int target) {
        int sum = 0;
        for (int& num : nums) {
            sum += num;
        }
        int diff = sum + target;
        if (diff < 0 || diff % 2 != 0) {
            return 0;
        }
        int n = nums.size(), neg = diff / 2;
        vector<vector<int>> dp(n + 1, vector<int>(neg + 1));
        dp[0][0] = 1;
        for (int i = 1; i <= n; i++) {
            int num = nums[i - 1];
            for (int j = 0; j <= neg; j++) {
                dp[i][j] = dp[i - 1][j];
                if (j >= num) {
                    dp[i][j] += dp[i - 1][j - num];
                }
            }
        }
        return dp[n][neg];
    }
};

 

 

 

回溯 ：

 

 

class Solution:
    def __init__(self):
        self.res = 0 
    def findTargetSumWays(self, nums: List[int], target: int) -> int:
        def backtracking(nums,target,cur,index):
            key = str(cur) + ',' + str(index)
            if key in memo:
                return memo[key]
            if index == len(nums):
                if cur == target:
                    return 1
                else:
                    return 0
            res = 0
            res += backtracking(nums,target,cur-nums[index],index+1)
            res += backtracking(nums,target,cur+nums[index],index+1)
            memo[key] = res
            return res
        memo = {}
        return backtracking(nums,target,0,0)

 

 

 

class Solution {
public:
    int res = 0;
    void dfs(vector<int>& nums, int i ,int cur_sum, int target) {
        if (i == nums.size()) {
            if (cur_sum == target) {
                res++;
            }
            return;
        }
        dfs(nums,i+1,cur_sum+nums[i],target);

        dfs(nums,i+1,cur_sum-nums[i],target);
    }
    int findTargetSumWays(vector<int>& nums, int target) {
      dfs(nums,0,0,target);
      return res;
    }
};