[Swift]LeetCode1157. 子数组中占绝大多数的元素 | Online Majority Element In Subarray

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Implementing the class MajorityChecker, which has the following API:

• MajorityChecker(int[] arr) constructs an instance of MajorityChecker with the given array arr;
• int query(int left, int right, int threshold) has arguments such that:
  • 0 <= left <= right < arr.length representing a subarray of arr;
  • 2 * threshold > right - left + 1, ie. the threshold is always a strict majority of the length of the subarray

Each query(...) returns the element in arr[left], arr[left+1], ..., arr[right] that occurs at least threshold times, or -1 if no such element exists.

 

Example:

MajorityChecker majorityChecker = new MajorityChecker([1,1,2,2,1,1]);
majorityChecker.query(0,5,4); // returns 1
majorityChecker.query(0,3,3); // returns -1
majorityChecker.query(2,3,2); // returns 2

 

Constraints:

• 1 <= arr.length <= 20000
• 1 <= arr[i] <= 20000
• For each query, 0 <= left <= right < len(arr)
• For each query, 2 * threshold > right - left + 1
• The number of queries is at most 10000

---

实现一个 MajorityChecker 的类，它应该具有下述几个 API：

• MajorityChecker(int[] arr) 会用给定的数组 arr 来构造一个 MajorityChecker 的实例。
• int query(int left, int right, int threshold) 有这么几个参数：
  • 0 <= left <= right < arr.length 表示数组 arr 的子数组的长度。
  • 2 * threshold > right - left + 1，也就是说阀值 threshold 始终比子序列长度的一半还要大。

每次查询 query(...) 会返回在 arr[left], arr[left+1], ..., arr[right] 中至少出现阀值次数 threshold 的元素，如果不存在这样的元素，就返回 -1。

 

示例：

MajorityChecker majorityChecker = new MajorityChecker([1,1,2,2,1,1]);
majorityChecker.query(0,5,4); // 返回 1
majorityChecker.query(0,3,3); // 返回 -1
majorityChecker.query(2,3,2); // 返回 2

 

提示：

• 1 <= arr.length <= 20000
• 1 <= arr[i] <= 20000
• 对于每次查询，0 <= left <= right < len(arr)
• 对于每次查询，2 * threshold > right - left + 1
• 查询次数最多为 10000

---

2304ms

class MajorityChecker {
    var arr:[Int] = [Int]()

    init(_ arr: [Int]) {
        self.arr = arr        
    }
    
    func query(_ left: Int, _ right: Int, _ threshold: Int) -> Int {
        var count:[Int] = [Int](repeating:0,count:200000)
        if left <= right
        {
            for i in left...right
            {
                if ++count[arr[i] - 1] == threshold
                {
                    return arr[i]
                }
            }
        }        
        return -1
    }
}

/*扩展Int类，实现自增++、自减--运算符*/
extension Int{
    //++前缀:先自增再执行表达示
    static prefix func ++(num:inout Int) -> Int {
        //输入输出参数num
        num += 1
        //返回加1后的数值
        return num
    }
}

/**
 * Your MajorityChecker object will be instantiated and called as such:
 * let obj = MajorityChecker(arr)
 * let ret_1: Int = obj.query(left, right, threshold)
 */

---

class MajorityChecker {
    var a:[Int] = [Int]()
    var big:[Int] = [Int]()
    var fs:[[Int]] = [[Int]]()
    var B:Int = 0

    init(_ arr: [Int]) {
        self.a = arr
        let n:Int = a.count
        self.B = Int(sqrt(Double(n)))
        var f:[Int] = [Int](repeating:0,count:20001)
        for v in a
        {
            f[v] += 1
        }
        var bigger:[Int] = [Int](repeating:0,count:n)
        var p:Int = 0
        for i in 1...20000
        {
            if f[i] >= B
            {                
                bigger[p] = i
                p += 1
            }
        }
        big = bigger[0..<p] 
        fs = [[Int]](repeating:[Int](repeating:0,count:n+1),count:p)
        for i in 0..<p
        {
            for j in 0..<n
            {
                if a[j] == big[i]
                {
                    fs[i][j+1] = fs[i][j] + 1
                }
                else
                {
                    fs[i][j+1] = fs[i][j]
                }
            }
        }
    }
    
    func query(_ left: Int, _ right: Int, _ threshold: Int) -> Int {
        if right - left + 1 <= 2*B
        {
            var best:Int = -1
            var f:Int = 0
            for i in  left...right
            {
                if best == a[i]
                {
                    f += 1

                }
                else if f == 0
                {
                    best = a[i]
                }
                else
                {
                    f -= 1
                }
            }
            var ct:Int = 0
            for i in  left...right
            {
                if a[i] == best
                {
                    ct += 1
                }
            }
            return ct >= threshold ? best : -1
        }
        else
        {
            for i in 0..<big.count
            {
                var ff:Int = fs[i][right+1] - fs[i][left]
                if ff >= threshold
                {
                    return big[i]
                }                
            }
            return -1
        }
    }
}

/**
 * Your MajorityChecker object will be instantiated and called as such:
 * let obj = MajorityChecker(arr)
 * let ret_1: Int = obj.query(left, right, threshold)
 */