2019-05-02 15:59:01 +08:00
|
|
|
|
# LeetCode 第 219 号问题：存在重复元素 II
|
|
|
|
|
|
|
|
|
|
|
|
> 本文首发于公众号「五分钟学算法」，是[图解 LeetCode ](<https://github.com/MisterBooo/LeetCodeAnimation>)系列文章之一。
|
|
|
|
|
|
>
|
|
|
|
|
|
> 个人网站：[https://www.cxyxiaowu.com](https://www.cxyxiaowu.com)
|
|
|
|
|
|
|
|
|
|
|
|
题目来源于 LeetCode 上第 219 号问题：存在重复元素 II。题目难度为 Easy，目前通过率为 34.8% 。
|
|
|
|
|
|
|
|
|
|
|
|
### 题目描述
|
|
|
|
|
|
|
|
|
|
|
|
给定一个整数数组和一个整数 *k*，判断数组中是否存在两个不同的索引 *i* 和 *j*，使得 **nums [i] = nums [j]**，并且 *i* 和 *j* 的差的绝对值最大为 *k*。
|
|
|
|
|
|
|
|
|
|
|
|
**示例 1:**
|
|
|
|
|
|
|
|
|
|
|
|
```
|
|
|
|
|
|
输入: nums = [1,2,3,1], k = 3
|
|
|
|
|
|
输出: true
|
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
|
|
**示例 2:**
|
|
|
|
|
|
|
|
|
|
|
|
```
|
|
|
|
|
|
输入: nums = [1,0,1,1], k = 1
|
|
|
|
|
|
输出: true
|
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
|
|
**示例 3:**
|
|
|
|
|
|
|
|
|
|
|
|
```
|
|
|
|
|
|
输入: nums = [1,2,3,1,2,3], k = 2
|
|
|
|
|
|
输出: false
|
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
|
|
### 题目解析
|
|
|
|
|
|
|
|
|
|
|
|
考虑用滑动窗口与查找表来解决。
|
|
|
|
|
|
|
|
|
|
|
|
* 设置查找表`record`，用来保存每次遍历时插入的元素，`record `的最大长度为`k `
|
|
|
|
|
|
* 遍历数组`nums`，每次遍历的时候在`record `查找是否存在相同的元素，如果存在则返回`true`，遍历结束
|
|
|
|
|
|
* 如果此次遍历在`record `未查找到，则将该元素插入到`record `中，而后查看`record `的长度是否为`k + 1`
|
|
|
|
|
|
* 如果此时`record `的长度是否为`k + 1`，则删减`record`的元素，该元素的值为`nums[i - k]`
|
|
|
|
|
|
* 如果遍历完整个数组`nums`未查找到则返回`false`
|
|
|
|
|
|
|
|
|
|
|
|
### 动画描述
|
|
|
|
|
|
|
2019-11-14 11:00:28 +08:00
|
|
|
|
|
2019-05-02 15:59:01 +08:00
|
|
|
|
|
|
|
|
|
|
### 代码实现
|
|
|
|
|
|
|
|
|
|
|
|
```
|
|
|
|
|
|
// 219. Contains Duplicate II
|
|
|
|
|
|
// https://leetcode.com/problems/contains-duplicate-ii/description/
|
|
|
|
|
|
// 时间复杂度: O(n)
|
|
|
|
|
|
// 空间复杂度: O(k)
|
|
|
|
|
|
class Solution {
|
|
|
|
|
|
public:
|
|
|
|
|
|
bool containsNearbyDuplicate(vector<int>& nums, int k) {
|
|
|
|
|
|
|
|
|
|
|
|
if(nums.size() <= 1) return false;
|
|
|
|
|
|
|
|
|
|
|
|
if(k <= 0) return false;
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
unordered_set<int> record;
|
|
|
|
|
|
for(int i = 0 ; i < nums.size() ; i ++){
|
|
|
|
|
|
|
|
|
|
|
|
if(record.find(nums[i]) != record.end()){
|
|
|
|
|
|
return true;
|
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
|
|
record.insert(nums[i]);
|
|
|
|
|
|
|
|
|
|
|
|
// 保持record中最多有k个元素
|
|
|
|
|
|
// 因为在下一次循环中会添加一个新元素,使得总共考虑k+1个元素
|
|
|
|
|
|
if(record.size() == k + 1){
|
|
|
|
|
|
record.erase(nums[i - k]);
|
|
|
|
|
|
}
|
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
|
|
return false;
|
|
|
|
|
|
}
|
|
|
|
|
|
};
|
|
|
|
|
|
```
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
2019-11-14 11:00:28 +08:00
|
|
|
|
|
