34-Search for a Range

【题目】　　

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].

【analyse】

1.由于对算法时间复杂度要求为O(log n)，故采用二分搜索的方法

2.由于需要查找的一个范围区间的数，有两种方式：1.根据二分查找到的第一数，然后左右遍历查找；2.采用递归的方式，查找出范围区间的边界值

【算法】

v1:
public int[] searchRange(int[] nums, int target) {
        int[] result=new int[2];
        int begin=-1;
        int end=-1;
        int left=0;
        int right=nums.length-1;
        int flag=-1;
        while(left<=right) {
            int mid=(left+right)/2;
            if(nums[mid]<target)
                left=mid+1;
            else if(nums[mid]>target) 
                right=mid-1;
            else {
                flag=mid;
                break;
            }
        }
        if(flag!=-1) {
            begin=flag;
            end=flag;
            while(begin>0) {
                if(nums[begin-1]==target)
                    begin--;
                else 
                    break;
            }
            while(end<nums.length-1) {
                if(nums[end+1]==target)
                    end++;
                else
                    break;
            }
        }
        result[0]=begin;
        result[1]=end;
        return result;
    }

v2:
public int[] searchRange(int[] nums,int target) {
　　int[] result=new int[2];
    result[0]=getBoundRange(nums,0,nums.length-1,target,true);
    result[1]=getBoundRange(nums,0,nums.length-1,target,false);
    return result;
}
public int getBoundRange(int[] nums,int left,int right,int target,boolean left) {
    if(left<=right) {
        int mid=(left+right)/2;
        if(nums[mid]==target) {
            if(left) {
                if(mid=0||nums[mid-1]<target) 
                      return mid;
                else 
                      return getBoundRange(nums,left,mid-1,target,left);
            }else {
                if(mid=nums.length-1||nums[mid+1]>target) 
                      return mid;
                else 
                      return getBoundRange(nums,mid+1,right,target,left);
            }
        }else if(nums[mid]<target) {
            return getBoundRange(nums,mid+1,right,target,left);
        }else {
            return getBoundRange(nums,left,mid-1,target,left);
        }
    }else {
        return -1;
    }    
}