README.md

259. 较小的三数之和

English Version

题目描述

给定一个长度为 n 的整数数组和一个目标值 target ，寻找能够使条件 nums[i] + nums[j] + nums[k] < target 成立的三元组  i, j, k 个数（0 <= i < j < k < n）。

 

示例 1：

输入: nums = [-2,0,1,3], target = 2
输出: 2 
解释: 因为一共有两个三元组满足累加和小于 2:
     [-2,0,1]
     [-2,0,3]

示例 2：

输入: nums = [], target = 0
输出: 0 

示例 3：

输入: nums = [0], target = 0
输出: 0 

 

提示:

• n == nums.length
• 0 <= n <= 3500
• -100 <= nums[i] <= 100
• -100 <= target <= 100

解法

双指针解决。

Python3

class Solution:
    def threeSumSmaller(self, nums: List[int], target: int) -> int:
        nums.sort()
        ans, n = 0, len(nums)
        for i in range(n):
            j, k = i + 1, n - 1
            while j < k:
                s = nums[i] + nums[j] + nums[k]
                if s >= target:
                    k -= 1
                else:
                    ans += k - j
                    j += 1
        return ans

Java

class Solution {
    public int threeSumSmaller(int[] nums, int target) {
        Arrays.sort(nums);
        int ans = 0;
        for (int i = 0, n = nums.length; i < n; ++i) {
            int j = i + 1;
            int k = n - 1;
            while (j < k) {
                int s = nums[i] + nums[j] + nums[k];
                if (s >= target) {
                    --k;
                } else {
                    ans += k - j;
                    ++j;
                }
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int threeSumSmaller(vector<int>& nums, int target) {
        sort(nums.begin(), nums.end());
        int ans = 0;
        for (int i = 0, n = nums.size(); i < n; ++i)
        {
            int j = i + 1, k = n - 1;
            while (j < k)
            {
                int s = nums[i] + nums[j] + nums[k];
                if (s >= target) --k;
                else
                {
                    ans += k - j;
                    ++j;
                }
            }
        }
        return ans;
    }
};

Go

func threeSumSmaller(nums []int, target int) int {
	sort.Ints(nums)
	ans := 0
	for i, n := 0, len(nums); i < n; i++ {
		j, k := i+1, n-1
		for j < k {
			s := nums[i] + nums[j] + nums[k]
			if s >= target {
				k--
			} else {
				ans += k - j
				j++
			}
		}
	}
	return ans
}

JavaScript

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var threeSumSmaller = function (nums, target) {
    nums.sort((a, b) => a - b);
    let ans = 0;
    for (let i = 0, n = nums.length; i < n; ++i) {
        let j = i + 1;
        let k = n - 1;
        while (j < k) {
            s = nums[i] + nums[j] + nums[k];
            if (s >= target) {
                --k;
            } else {
                ans += k - j;
                ++j;
            }
        }
    }
    return ans;
};

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