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：

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]

进阶：是否能在 O(n²) 的时间复杂度内解决？

解法

双指针解决。

Python3

class Solution:
    def threeSumSmaller(self, nums: List[int], target: int) -> int:
        def threeSumSmaller(nums, start, end, target):
            count = 0
            while start < end:
                if nums[start] + nums[end] < target:
                    count += (end - start)
                    start += 1
                else:
                    end -= 1
            return count

        nums.sort()
        n, count = len(nums), 0
        for i in range(n - 2):
            count += threeSumSmaller(nums, i + 1, n - 1, target - nums[i])
        return count

Java

class Solution {
    public int threeSumSmaller(int[] nums, int target) {
        Arrays.sort(nums);
        int n = nums.length;
        int count = 0;
        for (int i = 0; i < n - 2; ++i) {
            count += threeSumSmaller(nums, i + 1, n - 1, target - nums[i]);
        }
        return count;
    }

    private int threeSumSmaller(int[] nums, int start, int end, int target) {
        int count = 0;
        while (start < end) {
            if (nums[start] + nums[end] < target) {
                count += (end - start);
                ++start;
            } else {
                --end;
            }
        }
        return count;
    }
}

JavaScript

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var threeSumSmaller = function (nums, target) {
    let len = nums.length;
    if (len < 3) return 0;
    nums.sort((a, b) => a - b)
    let res = 0;
    for (let i = 0; i < len - 2; i++) {
        let left = i + 1, right = len - 1;
        if (nums[i] + nums[left] + nums[i + 2] >= target) break;
        while (left < right) {
            if (nums[i] + nums[left] + nums[right] < target) {
                res += (right - left);
                left++;
                continue;
            } else {
                right--;
                continue;
            }
        }
    }
    return res;
};

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