[Swift]LeetCode704. 二分查找 | Binary Search

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Given a sorted (in ascending order) integer array nums of nelements and a target value, write a function to search target in nums. If target exists, then return its index, otherwise return -1.

Example 1:

Input: nums = [-1,0,3,5,9,12], target = 9
Output: 4
Explanation: 9 exists in nums and its index is 4

Example 2:

Input: nums = [-1,0,3,5,9,12], target = 2
Output: -1
Explanation: 2 does not exist in nums so return -1 

Note:

1. You may assume that all elements in nums are unique.
2. n will be in the range [1, 10000].
3. The value of each element in nums will be in the range [-9999, 9999].

---

给定一个 n 个元素有序的（升序）整型数组 nums 和一个目标值 target  ，写一个函数搜索 nums 中的 target，如果目标值存在返回下标，否则返回 -1。

示例 1:

输入: nums = [-1,0,3,5,9,12], target = 9
输出: 4
解释: 9 出现在 nums 中并且下标为 4

示例 2:

输入: nums = [-1,0,3,5,9,12], target = 2
输出: -1
解释: 2 不存在 nums 中因此返回 -1 

提示：

1. 你可以假设 nums 中的所有元素是不重复的。
2. n 将在 [1, 10000]之间。
3. nums 的每个元素都将在 [-9999, 9999]之间。

---

292ms

class Solution {
    func search(_ nums: [Int], _ target: Int) -> Int {
        
        var left = 0
        var right = nums.count - 1
        var mid = (left + right) / 2
        
        while left <= right {
            if target == nums[mid] {
                return mid
            } else if target == nums[left] {
                return left
            } else if target == nums[right] {
                return right
            } else if target > nums[left] && target < nums[mid] {
                right = mid - 1
                mid = (left + mid) / 2
                left += 1
            } else if target > nums[mid] && target < nums[right] {
                left = mid + 1
                mid = (mid + right) / 2
                right -= 1
            } else {
                return -1
            } 
        }
        
        return -1
    }
}

---

Runtime: 296 ms

Memory Usage: 18.9 MB

class Solution {
    func search(_ nums: [Int], _ target: Int) -> Int {
        var low = 0
        var high = nums.count - 1
        var mid = (low + high) >> 1
        
        while low <= high {
            let val = nums[mid]
            if target == val {
                return mid
            } else if target < val {
                high = mid - 1
            } else {
                low = mid + 1 
            }
            mid = (low + high) >> 1
        }
        return -1
    }
}

---

312ms

class Solution {
    func search(_ nums: [Int], _ target: Int) -> Int {
        if nums.count == 1 {return nums[0] == target ? 0 : -1}
        else if nums.count == 0 {return -1}
        
        return binarySearch (nums, target, 0, nums.count-1)
    }
    
    func binarySearch(_ nums:[Int], _ target: Int, _ l: Int, _ r: Int) -> Int {
        if (r >= l) {
            var mid = l + (r-l)/2
            if nums[mid] == target {return mid}
            else if nums[mid] > target {return binarySearch(nums, target, l, mid-1) }
            else {return binarySearch(nums, target, mid+1, r)}
        }
        return -1
    }
}

---

352ms

class Solution {
    
    var start = 0;
    var end = 0;
    
    var center: Int {
        return (end + start) >> 1
    }
    
    func search(_ nums: [Int], _ target: Int) -> Int {
        end = nums.count
        while start != end - 1 {
            if nums[center] == target {
                return center
            }else if nums[center] < target  {
                start = center
            }else {
                end = center
            }
        }
        return nums[center] == target ? center : -1
    }
}

---

452ms

class Solution {
    func search(_ nums: [Int], _ target: Int) -> Int {
        func search(_ l: Int, _ r: Int) -> Int {
            if l == r && nums[l] != target {
                return -1
            }
            
            let mid = l + (r + 1 - l) / 2
            
            if nums[mid] == target {
                return mid
            } else if nums[mid] > target {
                return search(l, mid - 1)
            } else {
                return search(mid, r)
            }
        }
        
        return search(0, nums.count - 1)
    }
}