LeetCode 1150. Check If a Number Is Majority Element in a Sorted Array

原题链接在这里：https://leetcode.com/problems/check-if-a-number-is-majority-element-in-a-sorted-array/

题目：

Given an array nums sorted in non-decreasing order, and a number target, return True if and only if target is a majority element.

A majority element is an element that appears more than N/2 times in an array of length N.

Example 1:

Input: nums = [2,4,5,5,5,5,5,6,6], target = 5
Output: true
Explanation: 
The value 5 appears 5 times and the length of the array is 9.
Thus, 5 is a majority element because 5 > 9/2 is true.

Example 2:

Input: nums = [10,100,101,101], target = 101
Output: false
Explanation: 
The value 101 appears 2 times and the length of the array is 4.
Thus, 101 is not a majority element because 2 > 4/2 is false.

Note:

1. 1 <= nums.length <= 1000
2. 1 <= nums[i] <= 10^9
3. 1 <= target <= 10^9

题解：

Use binary search to find the first occurance of target. Make sure that first ovvurance index + n / 2 also points to target.

Time Complexity: O(logn).

Space: O(1).

AC Java:

 1 class Solution {
 2     public boolean isMajorityElement(int[] nums, int target) {
 3         if(nums == null || nums.length == 0){
 4             return false;
 5         }
 6         
 7         int n = nums.length;
 8         int l = 0;
 9         int r = n - 1;
10         while(l < r){
11             int mid = l + (r - l) / 2;
12             if(nums[mid] < target){
13                 l = mid + 1;
14             }else{
15                 r = mid;
16             }
17         }
18         
19         return l + n / 2 < n && nums[l + n / 2] == target;
20     }
21 }

类似Majority Element.