[LeetCode]Search for a Range

Description:
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].

题意理解： 其实这道题求target的起点和终点，并且要求时间复杂度为O(logn),所以可以通过用二分法来找到target的边界值。

code：

public int[] searchRange(int[] nums, int target) {    int[] result = new int[2];    result[0] = findFirst(nums, target);    result[1] = findLast(nums, target);    return result;}private int findFirst(int[] nums, int target){    int idx = -1;    int start = 0;    int end = nums.length - 1;    while(start <= end){        int mid = (start + end) / 2;        if(nums[mid] >= target){            end = mid - 1;        }else{            start = mid + 1;        }        if(nums[mid] == target) idx = mid;    }    return idx;}private int findLast(int[] nums, int target){    int idx = -1;    int start = 0;    int end = nums.length - 1;    while(start <= end){        int mid = (start + end) / 2;        if(nums[mid] <= target){            start = mid + 1;        }else{            end = mid - 1;        }        if(nums[mid] == target) idx = mid;    }    return idx; }