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1 | 1 | packagecom.fishercoder.solutions; |
2 | 2 |
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3 | | -/** |
4 | | - * 1150. Check If a Number Is Majority Element in a Sorted Array |
5 | | - * |
6 | | - * Given an array nums sorted in non-decreasing order, and a number target, return True if and only if target is a majority element. |
7 | | - * A majority element is an element that appears more than N/2 times in an array of length N. |
8 | | - * |
9 | | - * Example 1: |
10 | | - * Input: nums = [2,4,5,5,5,5,5,6,6], target = 5 |
11 | | - * Output: true |
12 | | - * Explanation: |
13 | | - * The value 5 appears 5 times and the length of the array is 9. |
14 | | - * Thus, 5 is a majority element because 5 > 9/2 is true. |
15 | | - * |
16 | | - * Example 2: |
17 | | - * Input: nums = [10,100,101,101], target = 101 |
18 | | - * Output: false |
19 | | - * Explanation: |
20 | | - * The value 101 appears 2 times and the length of the array is 4. |
21 | | - * Thus, 101 is not a majority element because 2 > 4/2 is false. |
22 | | - * |
23 | | - * Note: |
24 | | - * 1 <= nums.length <= 1000 |
25 | | - * 1 <= nums[i] <= 10^9 |
26 | | - * 1 <= target <= 10^9 |
27 | | - **/ |
28 | 3 | publicclass_1150 { |
29 | 4 | publicstaticclassSolution1 { |
30 | | -/**credit: https://leetcode.com/problems/check-if-a-number-is-majority-element-in-a-sorted-array/discuss/358130/Java-just-one-binary-search-O(logN))-0ms-beats-100*/ |
| 5 | +/** |
| 6 | + * credit: https://leetcode.com/problems/check-if-a-number-is-majority-element-in-a-sorted-array/discuss/358130/Java-just-one-binary-search-O(logN))-0ms-beats-100 |
| 7 | + */ |
31 | 8 | publicbooleanisMajorityElement(int[]nums,inttarget) { |
32 | 9 | intfirstIndex =findFirstOccur(nums,target); |
33 | 10 | intplusHalfIndex =firstIndex +nums.length /2; |
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