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  • 15. 3Sum

    package LeetCode_15
    
    /**
     * 15. 3Sum
     * https://leetcode.com/problems/3sum/
     *
     * Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0?
     * Find all unique triplets in the array which gives the sum of zero.
    Notice that the solution set must not contain duplicate triplets.
    
    Example 1:
    Input: nums = [-1,0,1,2,-1,-4]
    Output: [[-1,-1,2],[-1,0,1]]
    
    Example 2:
    Input: nums = []
    Output: []
    
    Example 3:
    Input: nums = [0]
    Output: []
    
    Constraints:
    1. 0 <= nums.length <= 3000
    2. -105 <= nums[i] <= 105
     * */
    class Solution {
        /*
        * solution: Two Pointer, Time:O(n^2), Space complexity:O(1)
        * */
        fun threeSum(nums: IntArray): List<List<Int>> {
            val result = ArrayList<ArrayList<Int>>()
            if (nums == null || nums.isEmpty()) {
                return result
            }
            nums.sort()
            var left = 0
            var right = 0
            for (i in nums.indices) {
                left = i + 1
                right = nums.size - 1
                //avoid duplicate
                if (i > 0 && nums[i] == nums[i - 1]) {
                    continue
                }
                while (left < right) {
                    val num = nums[i] + nums[left] + nums[right]
                    if (num == 0) {
                        //find out one result
                        result.add(arrayListOf(nums[i], nums[left], nums[right]))
                        //avoid duplicate
                        while (left < right && nums[left] == nums[left + 1]) {
                            left++
                        }
                        while (left < right && nums[right] == nums[right - 1]) {
                            right--
                        }
                        left++
                        right--
                    } else if (num < 0) {
                        //if sum less than 0, we need more large number, so left++
                        left++
                    } else {
                        right--
                    }
                }
            }
            return result
        }
    }
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  • 原文地址:https://www.cnblogs.com/johnnyzhao/p/13957184.html
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