从零开始的LC刷题(53): Contains Duplicate II 查找重复元素2

原题：

Given an array of integers and an integer k, find out whether there are two distinct indices i and j in the array such that nums[i] = nums[j] and the absolute difference between i and j is at most k.

Example 1:

Input: nums = [1,2,3,1], k = 3
Output: true

Example 2:

Input: nums = [1,0,1,1], k = 1
Output: true

Example 3:

Input: nums = [1,2,3,1,2,3], k = 2
Output: false

这个和上一个题相比麻烦就麻烦在要存储这个数上一次出现的位置，为此不得不看了看map的用法，结果：

Success

Runtime: 28 ms, faster than 93.62% of C++ online submissions for Contains Duplicate II.

Memory Usage: 15.4 MB, less than 24.76% of C++ online submissions for Contains Duplicate II.

代码：

class Solution {
public:
    bool containsNearbyDuplicate(vector& nums, int k) {
        unordered_map lib;
        for(int i=0;i

这次终于偷偷发现一个牛逼的结果，在details里找到的，手打的hash，给大家参考一下：

Success

Runtime: 20 ms, faster than 99.48% of C++ online submissions for Contains Duplicate II.

Memory Usage: 11.4 MB, less than 95.24% of C++ online submissions for Contains Duplicate II.

#define HASHSIZE 10009
#define NUMSIZE 100009
int hashTable[HASHSIZE];
long long getHash(long long n){
	if (n < 0) n = -n;
	return n% HASHSIZE;
}
struct node{
	int value;
	int next;
};

struct node arr[NUMSIZE];

class Solution {
public:
	bool containsNearbyDuplicate(vector& nums, int k) {
		int len = nums.size();

		int i,j,key;
		for (i = 0; i < HASHSIZE; i++)
			hashTable[i] = -1;
		
		for (i = 0; i < len; i++){
			key = getHash((long long)nums[i]);
			if (hashTable[key] < 0){
				hashTable[key] = i;
				arr[i].value = nums[i];
				arr[i].next = -1;
			}
			else{
				j = hashTable[key];
				while (j >= 0){
					if (nums[i] == arr[j].value && abs(i - j) <= k) return true;
					j = arr[j].next;
				}
				arr[i].next = hashTable[key];
				arr[i].value = nums[i];
				hashTable[key] = i;
			}
		}
		return false;
	}
};