LeetCode 1060. Missing Element in Sorted Array

本文转载自查看原文 2019-11-21 02:29 1147 Binary Search/ LeetCode

原题链接在这里：https://leetcode.com/problems/missing-element-in-sorted-array/

题目：

Given a sorted array A of unique numbers, find the K-th missing number starting from the leftmost number of the array.

Example 1:

Input: A = [4,7,9,10], K = 1
Output: 5 Explanation: The first missing number is 5.

Example 2:

Input: A = [4,7,9,10], K = 3
Output: 8 Explanation: The missing numbers are [5,6,8,...], hence the third missing number is 8.

Example 3:

Input: A = [1,2,4], K = 3
Output: 6 Explanation: The missing numbers are [3,5,6,7,...], hence the third missing number is 6.

Note:

1 <= A.length <= 50000
1 <= A[i] <= 1e7
1 <= K <= 1e8

题解：

If the missing numbers count of the whole array < k, then missing number must be after nums[n-1]. res = nums[n-1] + missingCount.

Otherwise, need to find out the starting index to calculate the missing number.

Use binary search to have mid as candidate.

If missing count < k, then must fall on the right side. l = mid + 1.

Time Complexity: O(logn). n = nums.length.

Space: O(1).

AC Java:

 1 class Solution {
 2     public int missingElement(int[] nums, int k) {
 3         int n = nums.length;
 4         if(nums[n - 1] - nums[0] - (n - 1 - 0) < k){
 5             return nums[n - 1] + k - missCount(nums, n - 1);
 6         }
 7         
 8         int l = 0;
 9         int r = n - 1;
10         
11         while(l < r){
12             int mid = l + (r - l) / 2;
13             if(missCount(nums, mid) < k){
14                 l = mid + 1;
15             }else{
16                 r = mid;
17             }
18         }
19         
20         return nums[l - 1] + k - missCount(nums, l - 1);
21     }
22     
23     private int missCount(int [] nums, int mid){
24         return nums[mid] - nums[0] - mid;
25     }
26 }

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