LeetCode 每日一题 494.目标和

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问题描述

简要思路

刚开始用暴力，显然超时了。
然后 使用 dfs 或者说是回溯优化，对每一个节点有 + 和 - 两个分支，在最终的叶子节点寻找 target。

代码

解法1（暴力）：

class Solution {
    
    
public:
    int mi(int x){
    
    
        int res = 2;
        for(int i = 1; i < x; i++){
    
    
            res *= 2;
        }
        return res;
    }

    int findTargetSumWays(vector<int>& nums, int target) {
    
    
        int n = nums.size();
        int m = mi(n + 1);
        vector<int> tree(m, 0);

        int ans = 0;
        for(int i = 0, j = 2; i < n; i++){
    
    
            int leaf = mi(i + 2);
            if(i == n - 1){
    
    
                while(j < leaf){
    
    
                    tree[j] = tree[j / 2] + nums[i];
                    if(tree[j] == target) ans++;
                    j++;
                    tree[j] = tree[j / 2] - nums[i];
                    if(tree[j] == target) ans++;
                    j++;
                }
            }
            while(j < leaf){
    
    
                tree[j] = tree[j / 2] + nums[i];
                j++;
                tree[j] = tree[j / 2] - nums[i];
                j++;
            }
        }

        return ans;
    }
};

解法2（dfs）：

class Solution {
    
    
public:
    int ans = 0;

    void dfs(vector<int>& nums, int deep, int sum, int target){
    
    
        if(deep == nums.size()) return;

        int sum1 = sum + nums[deep];
        int sum2 = sum - nums[deep];

        if(deep == nums.size() - 1){
    
    
            if(sum1 == target) ans++;
            if(sum2 == target) ans++;
        }

        dfs(nums, deep + 1, sum1, target);
        dfs(nums, deep + 1, sum2, target);
    }

    int findTargetSumWays(vector<int>& nums, int target){
    
    
        int n = nums.size();
        dfs(nums, 0, 0, target);
        return ans;
    }
};