LeetCode-----34. Search for a Range(查找范围)

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]

.
//[1,2,3,4,4,4，5,6] 4 [2,4]

改进的二分法log（n）：

public class Solution {
public int[] searchRange(int[] nums, int target) {
int start=0,mid,end=nums.length-1;
int [] arr={-1,-1};
while(start<end-1){
mid=(start+end)/2;
if(nums[mid]==target){end=mid;}
else if(nums[mid]<target){start=mid;}
else{end=mid;}
}

if(nums[start]==target){arr[0]=start;}
else if(nums[end]==target){arr[0]=end;}
else{arr[0]=-1;arr[1]=-1;return arr;}
start=0;end=nums.length-1;
while(start<end-1){
mid=(start+end)/2;
if(nums[mid]==target){start=mid;}
else if(nums[mid]<target){start=mid;}
else{end=mid;}
}
if(nums[end]==target){arr[1]=end;}
else if(nums[start]==target){arr[1]=start;}
else{arr[0]=-1;arr[1]=-1;return arr;}

return arr;
}
}