Given an array of integers and an integer k, find out whether there there are two distinct indices i and j in the array such that nums[i] = nums[j] and the difference between iand j is at most k.
思路分析:这题比较简单,可以直接定义一个长度最大为k的滑动窗口,用一个set维护窗口内的数字判断是否出现重复,使用两个指针start和end标记滑动窗口的两端,初始都是0,然后end不断进行扩展,扫描元素判断是否出现重复元素,直到发现end-start>k, 就开始移动start,并且在set中移除对应的元素。如果以为扫描到数组末尾还没有发现重复元素,那就可以返回false。时间复杂度和空间复杂度都是O(N)。
AC Code
public class Solution {
public boolean containsNearbyDuplicate(int[] nums, int k) {
//0842
Set<Integer> appearedNum = new HashSet<Integer>();
int start = 0, end = 0;
for(int i = 0; i < nums.length; i++){
if(!appearedNum.contains(nums[i])){
appearedNum.add(nums[i]);
end++;
} else return true;
if(end - start > k) {
appearedNum.remove(nums[start]);
start++;
}
}
return false;
//0848
}
}