lintcode462- Total Occurrence of Target- easy

Given a target number and an integer array sorted in ascending order. Find the total number of occurrences of target in the array.

Example

Given [1, 3, 3, 4, 5] and target = 3, return 2.

Given [2, 2, 3, 4, 6] and target = 4, return 1.

Given [1, 2, 3, 4, 5] and target = 6, return 0.

Challenge 

Time complexity in O(logn)

先二分find the first，再count次数。切记find the first的话，if(target == mid){end = mid;}

public class Solution {
    /*
     * @param A: A an integer array sorted in ascending order
     * @param target: An integer
     * @return: An integer
     */
    public int totalOccurrence(int[] A, int target) {
        // write your code here
        if (A == null || A.length == 0){
            return 0;
        }
        // find the first target
        int start = 0;
        int end = A.length - 1;
        while (start + 1 < end){
            int mid = start + (end - start) / 2;
            if (target <= A[mid]){
                end = mid;
            } else {
                start = mid;
            }
        }

        int firstIdx = 0;
        if (A[start] == target){
            firstIdx = start;
        } else if (A[end] == target){
            firstIdx = end;
        } else {
            return 0;
        }

        // count the occurance
        int count = 0;
        for (int i = firstIdx; i < A.length; i++){
            if (A[i] == target){
                count++;
            } else {
                break;
            }
        }

        return count;
    }
}