【leetcode】3Sum, 3Sum Closest

2024年6月9日 156次阅读

这两题一个套路。

先排个序；

外循环遍历数组，选出triplet第一个数；

内循环处理该数之后的子数组，分别从头向后、从尾向前遍历，找到triplet后两个数；

3Sum Closest须另设两个变量，分别保存最接近target的和，及其与target的差距，在内循环中更新；

时间代价O(n*n)。


3Sum

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) 
    {
        vector< vector<int> > result;
        sort(nums.begin(), nums.end());
        for (int i = 0; i < nums.size(); ++i)
        {
            if (nums[i] > 0)
                break;
            int front = i + 1;
            int back = nums.size() - 1;
            int target = -nums[i];
            while (front < back)
            {
                int sum = nums[front] + nums[back];
                if (sum < target) ++front;
                if (sum > target) --back;
                if (sum == target)
                {
                    vector<int> triplet(3, 0);
                    triplet[0] = nums[i];
                    triplet[1] = nums[front];
                    triplet[2] = nums[back];
                    result.push_back(triplet);
                    while (nums[front] == triplet[1] && front < back) ++front;
                    while (nums[back] == triplet[2] && front < back) --back;
                }
            }
            while (nums[i] == nums[i + 1] && i < nums.size() - 1) ++i;
        }
        return result;
    }
};

3Sum Closest

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target)
    {
        int result = 0;
        if (nums.size() < 3)
            return 0;
        sort(nums.begin(), nums.end());
        int closetDist = INT_MAX;
        for (int i = 0; i < nums.size(); ++i)
        {
            int left = i + 1;
            int right = nums.size() - 1;
            int sum = 0;
            while (left < right)
            {
                sum = nums[i] + nums[left] + nums[right];
                if (sum < target) ++left;
                if (sum > target) --right;
                if (sum == target) return sum;
                if (abs(sum - target) < closetDist)
                {
                    result = sum;
                    closetDist = abs(sum - target);
                }
            }
        }
        return result;
    }
};

Reference: 

https://leetcode.com/discuss/23595/share-my-solution-around-50ms-with-explanation-and-comments

点赞