LeetCode 34.Search for a Range

题目：

Given an array of integers sorted in ascending order, 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].

思路：

1.当数组长度为0时，返回[-1,-1];

2.当数组长度为1时，判断nums[0]与target关系;

3.当数组长度大于1时：

（1）因题目要求时间复杂度O(log n)，且整数数组以升序排列，故用二分法找出一个等于target的数组下标。

（2）起始坐标向前。

（3）结束坐标向后。

代码：

public class Solution {
    public int[] searchRange(int[] nums, int target) {
        int start = 0,end = nums.length-1,mid = 0;
        int resStart = -1,resEnd = -1;
        int[] resArr={-1,-1};
        if(nums.length == 0){
            resArr[0] = resArr[1] =-1;
            return resArr;
        }
        if(nums.length == 1){
            if(nums[0] == target){
                resArr[0] = resArr[1] =0;
                return resArr;
            }else{
                resStart = resEnd =-1;
                return resArr;
            }
        }
        while(start + 1 < end){
            mid = start + (end - start)/2;
            if(nums[mid] == target){
                resStart = resEnd = mid;
                break;
            }else if(nums[mid] > target){
                end = mid;
            }else{
                start = mid;
            }
        }
        if(resStart == -1 && nums[start] == target){
            resStart = resEnd = start;
        }
        if(resStart == -1 && nums[end] == target){
            resStart = resEnd = end;
        }
        while(resStart >0){
            if(nums[resStart-1] == target){
                resStart--;
            }else{
                break;
            }
        }
        while(resEnd < nums.length-1){
            if(nums[resEnd+1] == target){
                resEnd++;
            }else{
                break;
            }
        }
        resArr[0] = resStart;
        resArr[1] = resEnd;
        return resArr;
    }
}