LeetCode #18 4Sum

LeetCode #18 4Sum

Question

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note: The solution set must not contain duplicate quadruplets.

For example, given array S = [1, 0, -1, 0, -2, 2], and target = 0.
A solution set is:
[
  [-1,  0, 0, 1],
  [-2, -1, 1, 2],
  [-2,  0, 0, 2]
]

Solution

Approach #1

class Solution {
    func fourSum(_ nums: [Int], _ target: Int) -> [[Int]] {
        if nums.count < 4 { return [] }
        let sortedNums = nums.sorted()
        var results: [[Int]] = []
        for i in 0..<sortedNums.count - 3 {
            if i > 0, sortedNums[i] == sortedNums[i - 1] { continue }
            for j in i + 1..<sortedNums.count - 2 {
                if j > i + 1, sortedNums[j] == sortedNums[j - 1] { continue }
                var l = j + 1
                var r = sortedNums.count - 1
                while l < r {
                    let sum = sortedNums[i] + sortedNums[j] + sortedNums[l] + sortedNums[r]
                    if sum == target {
                        results.append([sortedNums[i], sortedNums[j], sortedNums[l], sortedNums[r]])
                        while l < r, sortedNums[l] == sortedNums[l + 1] { l += 1 }
                        l += 1
                        while l < r, sortedNums[r] == sortedNums[r - 1] { r -= 1 }
                        r -= 1
                    } else if sum < target {
                        while l < r, sortedNums[l] == sortedNums[l + 1] { l += 1 }
                        l += 1
                    } else {
                        while l < r, sortedNums[r] == sortedNums[r - 1] { r -= 1 }
                        r -= 1
                    }
                }
            }
        }
        return results
    }
}

Time complexity: O(n^3).

Space complexity: O(n).

转载请注明出处:http://www.cnblogs.com/silence-cnblogs/p/6893639.html

原文地址:https://www.cnblogs.com/silence-cnblogs/p/6893639.html