LeetCode 1150. Check If a Number Is Majority Element in a Sorted Array
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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 <= nums.length <= 1000
1 <= nums[i] <= 10^9
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 }
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