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[LeetCode] 259. 3Sum Smaller 三数之和较小值

 

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.

Example:

Input: nums = [-2,0,1,3], and target = 2
Output: 2 
Explanation: 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?

 
 
这道题是 3Sum 问题的一个变形,让我们求三数之和小于一个目标值,那么最简单的方法就是穷举法,将所有的可能的三个数字的组合都遍历一遍,比较三数之和跟目标值之间的大小,小于的话则结果自增1,参见代码如下:
 
解法一:
// O(n^3)
class Solution {
public:
    int threeSumSmaller(vector<int>& nums, int target) {
        int res = 0;
        sort(nums.begin(), nums.end());
        for (int i = 0; i < int(nums.size() - 2); ++i) {
            int left = i + 1, right = nums.size() - 1, sum = target - nums[i];
            for (int j = left; j <= right; ++j) {
                for (int k = j + 1; k <= right; ++k) {
                    if (nums[j] + nums[k] < sum) ++res;
                }
            }
        }
        return res;
    }
};

 

题目中的 Follow up 让我们在 O(n^2) 的时间复杂度内实现,那么借鉴之前那两道题 3Sum Closest 和 3Sum 中的方法,采用双指针来做,这里面有个 trick 就是当判断三个数之和小于目标值时,此时结果应该加上 right-left,因为数组排序了以后,如果加上 num[right] 小于目标值的话,那么加上一个更小的数必定也会小于目标值,然后将左指针右移一位,否则将右指针左移一位,参见代码如下:

 

解法二:

// O(n^2)
class Solution {
public:
    int threeSumSmaller(vector<int>& nums, int target) {
        if (nums.size() < 3) return 0;
        int res = 0, n = nums.size();
        sort(nums.begin(), nums.end());
        for (int i = 0; i < n - 2; ++i) {
            int left = i + 1, right = n - 1;
            while (left < right) {
                if (nums[i] + nums[left] + nums[right] < target) {
                    res += right - left;
                    ++left;
                } else {
                    --right;
                }
            }
        }
        return res;
    }
};

 

Github 同步地址:

https://github.com/grandyang/leetcode/issues/259

 

类似题目:

3Sum Closest

3Sum

Valid Triangle Number 

Two Sum Less Than K

 

参考资料:

https://leetcode.com/problems/3sum-smaller/

https://leetcode.com/problems/3sum-smaller/discuss/68817/Simple-and-easy-understanding-O(n2)-JAVA-solution

https://leetcode.com/problems/3sum-smaller/discuss/68820/Accepted-and-Simple-Java-O(n2)-solution-with-detailed-explanation

posted @ 2016-03-02 15:16  Grandyang  阅读(...)  评论(...编辑  收藏
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