[Swift]LeetCode315. 计算右侧小于当前元素的个数 | Count of Smaller Numbers After Self

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You are given an integer array nums and you have to return a new countsarray. The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i].

Example:

Input: [5,2,6,1]
Output: [2,1,1,0] 
Explanation:
To the right of 5 there are 2 smaller elements (2 and 1).
To the right of 2 there is only 1 smaller element (1).
To the right of 6 there is 1 smaller element (1).
To the right of 1 there is 0 smaller element.

---

给定一个整数数组 nums，按要求返回一个新数组 counts。数组 counts 有该性质： counts[i] 的值是  nums[i] 右侧小于 nums[i] 的元素的数量。

示例:

输入: [5,2,6,1]
输出: [2,1,1,0] 
解释:
5 的右侧有 2 个更小的元素 (2 和 1).
2 的右侧仅有 1 个更小的元素 (1).
6 的右侧有 1 个更小的元素 (1).
1 的右侧有 0 个更小的元素.

---

96ms

class Solution {
    func countSmaller(_ nums: [Int]) -> [Int] {
        var res: [Int] = [Int](repeating: 0, count: nums.count)
        var vals = nums.sorted()
        
        for i in 0..<nums.count {
            
            let index = binarySearch(vals, nums[i])
            res[i] = index
            vals.remove(at: index)
        }
        
        return res
    }
    
    func binarySearch(_ nums: [Int], _ val: Int) -> Int {
        
        var start = 0
        var end = nums.count-1
        var mid = (end - start) / 2
        
        while start < end {
            
            if nums[mid] >= val {
                end = mid
            }
            else {
                start = mid+1
            }
            mid = start + (end - start) / 2
        }
        
        return mid
    }
}

---

272ms

class Solution {
    func countSmaller(_ nums: [Int]) -> [Int] {
        var res = [Int]()
        
        var sorted = nums.sorted()
        
        func inedxOf(_ v : Int, _ arr : [Int]) -> Int {
            var l = 0
            var r = arr.count-1
            
            while l<=r {
                let mid = (l+r) >> 1
                if arr[mid] >= v {
                    r = mid - 1
                }else {
                    l = mid + 1
                }
            }
            
            return l
        }
        
        
        for i in 0..<nums.count {
            let c = nums[i]
            let index = inedxOf(c, sorted)
            res.append(index)
            sorted.remove(at: index)
        }
        
        return res
    }
}

---

2692ms

class Solution {
    func countSmaller(_ nums: [Int]) -> [Int] {
        
        var counts = Array(repeating:0, count: nums.count)
        
        for i in 0..<nums.count {
            
            var nos = 0
            
            for j in i+1..<nums.count {
                
                if nums[j] < nums[i] {
                    nos += 1
                }                
            }
            
            counts[i] = nos            
        }
        
        return counts        
    }
}

---

2924ms

class Solution {
    func countSmaller(_ nums: [Int]) -> [Int] {
        var results:[Int] = []
        for i in 0..<nums.count {
            var smaller = 0
            for j in i..<nums.count {
                if nums[j] < nums[i] {
                    smaller += 1
                }
            }
            results.append(smaller)
        }
        return results
    }
}

---

4168ms

class Solution {
    func countSmaller(_ nums: [Int]) -> [Int] {
 var res = [Int].init()
    guard nums.count>0 else {
       return res
    }
    if nums.count == 1 {
        return [0]
    }
    for i in 0..<nums.count-1 {
        var count = 0
        
        for j in (i+1)..<nums.count{
            if nums[j] < nums[i]{
                count += 1
            }
        }
        res.append(count)
    }
    res.append(0)
    return res
    }
}