leetcode 220: Contains Duplicate III

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Contains Duplicate III

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

[思路]

维持一个长度为k的window, 每次检查新的值是否与原来窗口中的所有值的差值有小于等于t的. 如果用两个for循环会超时O(nk). 使用treeset( backed by binary search tree) 的subSet函数,可以快速搜索. 复杂度为 O(n logk)

[CODE]

import java.util.SortedSet;

public class Solution {
    public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
        //input check
        if(k<1 || t<0 || nums==null || nums.length<2) return false;
        
        SortedSet<Long> set = new TreeSet<Long>();
        
        for(int j=0; j<nums.length; j++) {
            SortedSet<Long> subSet =  set.subSet((long)nums[j]-t, (long)nums[j]+t+1);
            if(!subSet.isEmpty()) return true;
            
            if(j>=k) {
                set.remove((long)nums[j-k]);
            }
            set.add((long)nums[j]);
        }
        return false;
    }
}

leetcode 220: Contains Duplicate III

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原文地址：http://blog.csdn.net/xudli/article/details/46323247