天天看点

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
    }
}