Search for a Range

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

Your algorithm's runtime complexity must be in the order of O(log n).

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

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

A solution of o(logn). Do two runs of binary search for the index of largest element strictly less than target, and the index of largest element strictly less than target+1. The range is between this two indices. 

public class Solution {
    public int[] searchRange(int[] A, int target) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(A==null || A.length==0)return null;
        int[] res = new int[2];
        res[0]=searchMaxLessThan(A,target,0,A.length-1);
        res[1]=searchMaxLessThan(A,target+1,0,A.length-1);
        if(res[0]==res[1]){
            res[0]=-1;
            res[1]=-1;
        }else{
            res[0]++;
        }
        return res;
    }
    
    public int searchMaxLessThan(int[] A, int target, int start, int end){
        
        if(start==end) return A[start]<target?start:start-1;
        if(start==end-1) return A[end]<target?end:(A[start]<target?start:start-1);
        int mid = (start+end)/2;
        if(A[mid]>=target){
            end=mid-1;
        }else{
            start=mid;
        }
        return searchMaxLessThan(A,target,start,end);
    }
}

 第二遍：

很巧妙的采用两次循环来做。

public class Solution {
    public int[] searchRange(int[] A, int target) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int start = 0;
        int end = A.length - 1;
        int l = -1;
        int r = -1;
        while(start <= end){
            int mid = (start + end) / 2;
            if(A[mid] == target) l = mid;
            if(target <= A[mid]) end = mid - 1;
            else start = mid + 1;
        }
        start = 0;
        end = A.length - 1;
        while(start <= end){
            int mid = (start + end) / 2;
            if(A[mid] == target) r = mid;
            if(target >= A[mid]) start = mid + 1;
            else end = mid - 1;
        }
        return new int[]{l, r};
    }
}

 第三遍：

public class Solution {
    public int[] searchRange(int[] A, int target) {
        int len = A.length;
        int srt = 0, end = len - 1;
        int left = -1, right = -1;
        while(srt <= end){
            int mid = (srt + end) / 2;
            if(A[mid] < target) srt = mid + 1;
            else if(A[mid] > target) end = mid - 1;
            else{
                left = mid;
                end = mid - 1;
            }
        }
        srt = 0; end = len - 1;
        while(srt <= end){
            int mid = (srt + end) / 2;
            if(A[mid] < target) srt = mid + 1;
            else if(A[mid] > target) end = mid - 1;
            else{
                right = mid;
                srt = mid + 1;
            }
        }
        return new int[]{left, right};
    }
}