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

Given an array of integers nums 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].

Example 1:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

用两次binary search，第一次找到左边界l，第二次binary search在[l, r] 中找右边界，注意要reset r = nums.length

time: O(logN), space: O(1)

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int[] res = {-1, -1};
        int l = 0, r = nums.length - 1;
        if(nums == null || nums.length == 0) return res;
        while(l < r) {
            int m = l + (r - l) / 2;
            if(nums[m] < target)
                l = m + 1;
            else
                r = m;
        }
        if(nums[l] != target) return res;
        res[0] = l;
        
        r = nums.length;
        while(l < r) {
            int m = l + (r - l) / 2;
            if(nums[m] > target)
                r = m;
            else
                l = m + 1;
        }
        res[1] = r - 1;
        return res;
    }
}

二刷：

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

三刷：

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