LeetCode 34. Find First and Last Position of Element in Sorted Array

原题链接在这里：https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/description/

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].

题解：

是Search Insert Position的变形，就是找边界情况。

只需要做两遍Binary Search.

第一遍用来找左边界，若是取mid小于target, 左边界会在后半部分. 若mid 大于 或者 等于 target时左边界都会在前半部分。

第二遍用来找右边界，原理相同。

然后如何判断是否找到过target呢，若是找到了target, 那么ll 不会在rr 的右侧，若是如此更改res结果，否则直接返回res.

Time Complexity: O(logn). Space: O(1).

public class Solution {
    public int[] searchRange(int[] nums, int target) {
        //Method 2
        int [] res = {-1,-1};
        if(nums == null || nums.length == 0){
            return res;
        }
        //find left bound
        int ll = 0;
        int lr = nums.length - 1;
        while(ll<=lr){
            int mid = ll+(lr-ll)/2;
            if(nums[mid] < target){
                ll = mid+1;
            }else{
                lr = mid-1;
            }
        }
        //find right bound
        int rl = 0;
        int rr = nums.length - 1;
        while(rl<=rr){
            int mid = rl+(rr-rl)/2;
            if(nums[mid] <= target){
                rl = mid+1;
            }else{
                rr = mid-1;
            }
        }
        //check if target is found
        if(ll>rr){
            return res;
        }
        res[0] = ll;
        res[1] = rr;
        return res;
    }
}

 类似First Bad Version, Find K Closest Elements.