[Swift]LeetCode300. 最长上升子序列 | Longest Increasing Subsequence

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Given an unsorted array of integers, find the length of longest increasing subsequence.

Example:

[10,9,2,5,3,7,101,18]
[2,3,7,101]
4

Note:

• There may be more than one LIS combination, it is only necessary for you to return the length.
• Your algorithm should run in O(n2) complexity.

Follow up: Could you improve it to O(n log n) time complexity?

---

给定一个无序的整数数组，找到其中最长上升子序列的长度。

示例:

[10,9,2,5,3,7,101,18]
[2,3,7,101]，
4

说明:

• 可能会有多种最长上升子序列的组合，你只需要输出对应的长度即可。
• 你算法的时间复杂度应该为 O(n2) 。

进阶: 你能将算法的时间复杂度降低到 O(n log n) 吗?

---

16ms

class Solution {
    func lengthOfLIS(_ nums: [Int]) -> Int {
        if nums.count < 2 {
            return nums.count
        }
        var res = Array(repeating: 0, count: nums.count)
        res[0] = nums[0]
        var count = 1
        var j = 0
        for i in 0..<nums.count {
            let tmp = nums[i]
            if tmp > res[count - 1] {
                j = count
                count += 1
            }else {
                var left = 0
                var right = count - 1
                while left < right {
                    let mid = (left + right) / 2
                    if res[mid] >= nums[i] {
                        right = mid
                    }else {
                        left = mid + 1
                    }
                }
                j = left
            }
            res[j] = nums[i]
        }
        return count
    }
}

---

56ms

class Solution {
    func lengthOfLIS(_ nums: [Int]) -> Int {
        var results = [Int]()
        for num in nums {
            var needAdd = true
            if results.count > 0 {
                for index in 0..<results.count {
                    if num <= results[index] {
                        results[index] = num
                        needAdd = false
                        break
                    }
                }
            }
            if needAdd {
                results.append(num)
            }
        }
        return results.count
    }
}

---

300ms

class Solution {
    func lengthOfLIS(_ nums: [Int]) -> Int {
        guard !nums.isEmpty else {
            return 0
        }
        
        var dps = Array(repeating: 1, count: nums.count)
        var result = 1
        for index in 1..<nums.count {
            let num = nums[index]
            for j in 0..<index {
                if num > nums[j] {
                    let length = dps[j] + 1
                    if length > dps[index] {
                        dps[index] = length
                        result = max(result, dps[index])
                    }
                }
            }
        }

        return result
    }
}

---

312ms

class Solution {
    func lengthOfLIS(_ nums: [Int]) -> Int {
         var length_global = 0
    var length_current = [Int].init(repeating:1 , count: nums.count)
    
    for i in 0..<nums.count {
        for j in 0..<i {
            if nums[i] > nums[j] {
                length_current[i] = max(length_current[i], length_current[j] + 1)
            }
        }
        length_global = max(length_global, length_current[i])
    }
    
    return length_global
    }
}

---

376ms

class Solution {
    func lengthOfLIS(_ nums: [Int]) -> Int {
        guard !nums.isEmpty else {
            return 0
        }
        
        var dps = Array(repeating: 1, count: nums.count)
        var result = 1
        for index in 1..<nums.count {
            let num = nums[index]
            for j in 0..<index where num > nums[j] {
                let length = dps[j] + 1
                dps[index] = max(length, dps[index])
                result = max(result, dps[index])
            }
        }

        return result
    }
}

---

388ms

class Solution {
    func lengthOfLIS(_ nums: [Int]) -> Int {
        
        if nums.count == 0 {
            return 0
        }
        
        var result = 1
        var dArr = [Int](repeating: 1, count: nums.count)
        
        for i in 0..<nums.count {

            for j in 0..<i {
                
                if(nums[j] < nums[i] && (dArr[j] + 1 >= dArr[i])) {
                    dArr[i] = dArr[j] + 1
                }
            }
            result = dArr[i] > result ? dArr[i] : result
        }
        return result
    }
}