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【Leetcode】Contains Duplicate II


题目链接:https://leetcode.com/problems/contains-duplicate-ii/

题目:

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 difference between i and j is at most k.

思路:

1、用HashMap保存元素的下标,如果出现重复元素,则判断是否满足 i-j<=k,不满足则更新HashMap保存的下标,因为后面的元素可能还有重复,要跟最近的重复元素比较距离。

2、用Set,具体看代码。。。时间久了我也忘了具体怎么做的。。。。
算法1:

public boolean containsNearbyDuplicate(int[] nums, int k) {
		HashMap<Integer, Integer> set = new HashMap<Integer, Integer>();
		for (int i = 0; i < nums.length; i++) {
			if (!set.containsKey(nums[i])) {
				set.put(nums[i], i);// 记录起始位置
			} else {
				if (i - set.get(nums[i]) <= k) {
					return true;
				} else {
					set.put(nums[i], i);
				}
			}
		}
		return false;
	}




算法2:

public boolean containsNearbyDuplicate(int[] nums, int k) {
		Set<Integer> set = new HashSet<Integer>();
		int start = 0, end = 0;
		for (int i = 0; i < nums.length; i++) {
			if (!set.contains(nums[i])) {
				set.add(nums[i]);
				end++;
			} else
				return true;
			if (end - start > k) {
				set.remove(nums[start]);
				start++;
			}
		}
		return false;
	}




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