Search for a Range leetcode

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].

这道题用两次二分法分别确定左边界和右边界

时间复杂度O(logn) 空间复杂度是O(1)

public class Solution {
    public int[] searchRange(int[] nums, int target) {
        int[] res = {-1, -1};
        if (nums == null || nums.length == 0) {
            return res;
        }
        int left = 0;
        int right = nums.length - 1;
        while (left <= right) {
            int mid = (left + right) / 2;
            if (target > nums[mid]) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        int start = 0;
        int end = nums.length - 1;
        while (start <= end) {
            int mid = (start + end) / 2;
            if (target >= nums[mid]) {
                start= mid + 1;
            } else {
                end = mid - 1;
            }
        }
        if (left <= end) {
            res[0] = left;
            res[1] = end;
        }
        return res;
    }
}

public class Solution {
    public int[] searchRange(int[] A, int target) {
        
        int [] result = {-1,-1};
        if (A.length == 0){
            return result;
        }
        int str = 0;
        int end = A.length - 1;
        int mid;
        // search for left bound
        while (str + 1 < end){
            mid = str + (end - str) / 2;
            if (A[mid] < target){
                str = mid;
            } else if (A[mid] == target){
                end = mid;
            } else {
                end = mid;
            }
        }
        if (A[str] == target){
            result[0] = str;
        } else if (A[end] == target){
            result[0] = end;
        } else {
            result[0] = result[1] = -1;
            return result;
        }
        // search for right bound
        str = 0;
        end = A.length - 1;
        while (str + 1 < end){
            mid = str + (end - str) / 2;
            if (A[mid] < target){
                str = mid;
            } else if (A[mid] == target){
                str = mid;
            } else {
                end = mid;
            }
        }
        if (A[end] == target){
            result[1] = end;
        } else if (A[str] == target){
            result[1] = str;
        } else {
            result[0] = result[1] = -1;
            return result;
        }
        return result;
    }
}