Search for a Range 61

Question

Given a sorted array of n integers, find the starting and ending position of a given target value.

If the target is not found in the array, return [-1, -1].

Example

Given [5, 7, 7, 8, 8, 10] and target value 8,

return [3, 4].

Challenge

O(log n) time.

Solution

做两遍binary search。第一遍找第一个出现的target，第二遍找最后出现的target。

代码如下：

public class Solution {
    /** 
     *@param A : an integer sorted array
     *@param target :  an integer to be inserted
     *return : a list of length 2, [index1, index2]
     */
    public int[] searchRange(int[] A, int target) {
        // write your code here
        if (A == null || A.length == 0){
            return new int[]{-1, -1};
        }

        int start = 0;
        int end = A.length - 1;
        int[] bound = new int[2];

        while (start + 1 < end){
            int mid = start + (end - start)/2;
            if (A[mid] == target){
                end = mid;
            }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;
        }

        start = 0; end = A.length - 1;

        while (start + 1 < end){
            int mid = start + (end - start)/2;
            if (A[mid] == target){
                start = mid;
            }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;
    }
}