[Swift]LeetCode1150. 检查一个数是否在数组中占绝大多数 | Check If a Number Is Majority Element in a Sorted Array...

た入场券 2021-10-29 08:22 95阅读 0赞

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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`

--------------------

给出一个按 非递减 顺序排列的数组 `nums`，和一个目标数值 `target`。假如数组 `nums` 中绝大多数元素的数值都等于 `target`，则返回 `True`，否则请返回 `False`。

所谓占绝大多数，是指在长度为 `N` 的数组中出现必须 超过 `N/2` 次。

示例 1：

输入：nums = [2,4,5,5,5,5,5,6,6], target = 5
    输出：true
    解释：
    数字 5 出现了 5 次，而数组的长度为 9。
    所以，5 在数组中占绝大多数，因为 5 次 > 9/2。

示例 2：

输入：nums = [10,100,101,101], target = 101
    输出：false
    解释：
    数字 101 出现了 2 次，而数组的长度是 4。
    所以，101 不是 数组占绝大多数的元素，因为 2 次 = 4/2。

提示：

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

--------------------

28 ms

1 class Solution {
     2     func isMajorityElement(_ nums: [Int], _ target: Int) -> Bool {
     3         var cnt:Int = 0
     4         for x in nums
     5         {
     6             if x == target
     7             {
     8                 cnt += 1
     9             }
    10         }
    11         return cnt > nums.count/2        
    12     }
    13 }

转载于:https://www.cnblogs.com/strengthen/p/11333853.html

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