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 occurance index + n / 2 also points to target.

Why can't directly check if nums[n / 2] == target. n = nums.length. e.g.[0, 1, 100, 100]. nums[n / 2] = 100. The occurance of 100 is 2, 2 is not > 4/2. The result should be false. 

But if check nums[n / 2] == target, then it returns true. 

Time Complexity: O(logn).

Space: O(1).

AC Java:

class Solution {
    public boolean isMajorityElement(int[] nums, int target) {
        if(nums == null || nums.length == 0){
            return false;
        }
        
        int n = nums.length;
        int l = 0;
        int r = n - 1;
        while(l < r){
            int mid = l + (r - l) / 2;
            if(nums[mid] < target){
                l = mid + 1;
            }else{
                r = mid;
            }
        }
        
        return l + n / 2 < n && nums[l + n / 2] == target;
    }
}

类似Majority Element.