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].

Solution1：

思路：
兩次二分查找分別查找target的start_index和end_index,
為處理重復(fù)target在nums中的情況，將nums[mid]==target的情況specify：
(1)在二分查找start_index時(shí)，當(dāng)nums[mid]==target時(shí): j = mid使得能夠繼續(xù)向左查找到target的起點(diǎn)
(1)在二分查找end_index時(shí)，當(dāng)nums[mid]==target時(shí): i = mid使得能夠繼續(xù)向右查找到target的起點(diǎn)，但因?yàn)閙id is biased to the left when mid = (i + j) / 2 比如[0, 1, 2, 3] mid = 1 但其實(shí)應(yīng)該是1.5，這樣因?yàn)槠蜃笏詴?huì)導(dǎo)致getting stuch when 比如 [7, 8]找7，mid無法到8。所以需要當(dāng)找end_index時(shí)要向右biased, 即int mid = (i + j + 1) / 2

Time Complexity: O(logN) Space Complexity: O(1)

Solution2：BinarySearch Template

思路：
Time Complexity: O(logN) Space Complexity: O(1)

Solution1 Code：

class Solution {
    public int[] searchRange(int[] nums, int target) {
        if(nums == null || nums.length == 0) return new int[]{-1, -1};
        if(nums.length == 1) {
            return nums[0] == target ? new int[]{0, 0} : new int[]{-1, -1};
        }
        
        int i = 0, j = nums.length - 1;
        int ret[] = new int[]{-1, -1};
        
        // Search for the left one
        while (i < j) { 
            int mid = (i + j) / 2;
            if(nums[mid] < target) i = mid + 1;
            else if(nums[mid] > target) j = mid - 1;
            else j = mid;
        }
        if (nums[i] != target) return ret;
        else ret[0] = i;

        // Search for the right one
        j = nums.length - 1; 
        while(i < j) {
            int mid = (i + j + 1) / 2;    // Made mid biased to the right
            if(nums[mid] < target) i = mid + 1;
            else if(nums[mid] > target) j = mid - 1;
            else i = mid;   
        }
        ret[1] = j;   
        return ret;
    }
}

Solution2 BinarySearch Template Code：

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