[Leetcode]3Sum Smaller

最新推荐文章于 2021-03-22 20:26:31 发布

dyllanzhou

最新推荐文章于 2021-03-22 20:26:31 发布

阅读量449

点赞数

CC 4.0 BY-SA版权

分类专栏： Two Pointers Array 文章标签： leetcode 算法

本文链接：https://blog.youkuaiyun.com/dyllanzhou/article/details/49408041

Array 同时被 2 个专栏收录

35 篇文章

订阅专栏

Two Pointers

14 篇文章

订阅专栏

本文介绍了一种算法，用于解决寻找数组中三个数相加小于特定目标值的问题，并提供了一个示例代码实现，该算法的时间复杂度为O(n²)。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given an array of n integers nums and a target, find the number of index triplets

i,
 j, k

with 0 <= i < j < k < n that satisfy the condition

nums[i]
 + nums[j] + nums[k] < target

For example, given nums = [-2, 0, 1, 3], and target = 2.

Return 2. Because there are two triplets which sums are less than 2:

[-2, 0, 1]
[-2, 0, 3]

Follow up:
Could you solve it in O(n2) runtime?

int threeSumSmaller(vector<int>&nums, int target){
    int size = nums.size();
    sort(nums.begin(),nums.end());
    int cnt = 0;
    for(int i = 0;i < size-2;i++){
        int j = i+1,k = size - 1;
        while(j < k){
           int sum = nums[i] + nums[j] + nums[k];           
           if(sum < target)cnt++;
           k--;  
        }         
    }
    return cnt;
}