Count the number of occurrences in a sorted array

Posted 2020-06-07

Given a sorted array arr[] and a number x, write a function that counts the occurrences of x in arr[]. Expected time complexity is O(Logn)

Examples:

  Input: arr[] = {1, 1, 2, 2, 2, 2, 3,},   x = 2
  Output: 4 // x (or 2) occurs 4 times in arr[]

  Input: arr[] = {1, 1, 2, 2, 2, 2, 3,},   x = 3
  Output: 1 

  Input: arr[] = {1, 1, 2, 2, 2, 2, 3,},   x = 1
  Output: 2 

  Input: arr[] = {1, 1, 2, 2, 2, 2, 3,},   x = 4
  Output: -1 // 4 doesn‘t occur in arr[] 

Method 1 (Linear Search)
Linearly search for x, count the occurrences of x and return the count.

Time Complexity: O(n)

Method 2 (Use Binary Search)
1) Use Binary search to get index of the first occurrence of x in arr[]. Let the index of the first occurrence be i.
2) Use Binary search to get index of the last occurrence of x in arr[]. Let the index of the last occurrence be j.
3) Return (j – i + 1);

 

就是search for a range的变种，

search for a range的代码：

/**
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 */

public class Solution {
    public int[] searchRange(int[] A, int target) {
        int start, end, mid;
        int[] bound = new int[2]; 
        
        // search for left bound
        start = 0; 
        end = A.length - 1;
        while (start + 1 < end) {
            mid = start + (end - start) / 2;
            if (A[mid] == target) {
                end = mid;     //让右指针right（也就是这里的end）往左边靠拢，也就是找到第一个等于target的数。
            } else if (A[mid] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        if (A[start] == target) {
            bound[0] = start;
        } else if (A[end] == target) {
            bound[0] = end;
        } else {
            bound[0] = bound[1] = -1;
            return bound;
        }
        
        // search for right bound
        start = 0;
        end = A.length - 1;
        while (start + 1 < end) {
            mid = start + (end - start) / 2;
            if (A[mid] == target) {
                start = mid;  //让左指针left（也就是这里的start）往右边靠拢，直到start+1<end，找到最后一个等于target的数。
            } else if (A[mid] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        if (A[end] == target) {
            bound[1] = end;
        } else if (A[start] == target) {
            bound[1] = start;
        } else {
            bound[0] = bound[1] = -1;
            return bound;
        }
        
        return bound;
    }
}