【LetMeFly】1287.有序数组中出现次数超过25%的元素:遍历
力扣题目链接:https://leetcode.cn/problems/element-appearing-more-than-25-in-sorted-array/
给你一个非递减的 有序整数数组,已知这个数组中恰好有一个整数,它的出现次数超过数组元素总数的 25%。
请你找到并返回这个整数
示例:
输入:arr = [1,2,2,6,6,6,6,7,10]
输出:6
提示:
1 <= arr.length <= 10^4
0 <= arr[i] <= 10^5
解题方法:遍历
对于长度为 n n n的有序 数组,对于一个首次出现下标为 i i i的数:
整个数出现次数超过了 25 % 25\% 25% 等价于 下标 ⌊ n 4 ⌋ + i \lfloor\frac{n}{4}\rfloor+i ⌊4n⌋+i仍为这个数。
- 时间复杂度 O ( l e n ( a r r ) ) O(len(arr)) O(len(arr))
- 空间复杂度 O ( 1 ) O(1) O(1)
AC代码
C++
cpp
/*
* @Author: LetMeFly
* @Date: 2025-02-17 14:38:51
* @LastEditors: LetMeFly.xyz
* @LastEditTime: 2025-02-17 14:42:26
*/
class Solution {
public:
int findSpecialInteger(vector<int>& arr) {
int locDiff = arr.size() / 4;
for (int i = 0; 1; i++) {
if (arr[i] == arr[i + locDiff]) {
return arr[i];
}
}
}
};
Python
python
'''
Author: LetMeFly
Date: 2025-02-17 14:38:57
LastEditors: LetMeFly.xyz
LastEditTime: 2025-02-17 14:43:22
'''
from typing import List
class Solution:
def findSpecialInteger(self, arr: List[int]) -> int:
locDiff = len(arr) // 4
for i in range(len(arr)):
if arr[i] == arr[i + locDiff]:
return arr[i]
return -1 # Fake Return
Java
java
/*
* @Author: LetMeFly
* @Date: 2025-02-17 14:39:05
* @LastEditors: LetMeFly.xyz
* @LastEditTime: 2025-02-17 14:44:38
*/
class Solution {
public int findSpecialInteger(int[] arr) {
int locDiff = arr.length / 4;
for (int i = 0; true; i++) {
if (arr[i] == arr[i + locDiff]) {
return arr[i];
}
}
}
}
Go
go
/*
* @Author: LetMeFly
* @Date: 2025-02-17 14:39:01
* @LastEditors: LetMeFly.xyz
* @LastEditTime: 2025-02-17 14:47:00
*/
package main
func findSpecialInteger(arr []int) int {
locDiff := len(arr) / 4
for i := 0; true; i++ {
if arr[i] == arr[i + locDiff] {
return arr[i]
}
}
return -1 // Fake Return - 及时上面是for true也需要return
}
同步发文于CSDN和我的个人博客,原创不易,转载经作者同意后请附上原文链接哦~
千篇源码题解已开源
Tisfy:https://blog.letmefly.xyz/2025/02/17/LeetCode 1287.有序数组中出现次数超过25的元素