leetcode 220. Contains Duplicate III

最新推荐文章于 2024-10-16 14:07:42 发布

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最新推荐文章于 2024-10-16 14:07:42 发布

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本文链接：https://blog.youkuaiyun.com/qq_27262609/article/details/70233907

本文介绍了一种使用TreeMap和HashMap解决寻找数组中是否存在两个不同的索引i和j的方法，使得nums[i]和nums[j]之间的绝对差小于等于t且i和j之间的绝对差小于等于k。

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

public class Solution {
    public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
        TreeMap<Integer,Integer> map=new TreeMap<>();
        HashMap<Integer,Integer> helpmap=new HashMap<>();
        if(t<0)return false;
        for(int i=0;i<=k&&i<nums.length;i++){
            if(map.containsKey(nums[i])){
                return true;
            }
            map.put(nums[i],i);

            if(map.lowerKey(nums[i])!=null&&nums[i]-map.lowerKey(nums[i])<=t||map.higherKey(nums[i])!=null&&map.higherKey(nums[i])-nums[i]<=t){
                if(map.lowerKey(nums[i])!=null){
                    if(nums[i]-map.lowerKey(nums[i])<0)continue;
                }else if(map.higherKey(nums[i])!=null){
                    if(map.higherKey(nums[i])-nums[i]<0)continue;
                }
                return true;
            }
            helpmap.put(i,nums[i]);
        }
        for(int i=k+1;i<nums.length;i++){
            int temp=helpmap.get(i-k-1);
            map.remove(temp);
            if(map.containsKey(nums[i])){
                return true;
            }
            map.put(nums[i],i);
            if(map.lowerKey(nums[i])!=null&&nums[i]-map.lowerKey(nums[i])<=t||map.higherKey(nums[i])!=null&&map.higherKey(nums[i])-nums[i]<=t){
                return true;
            }
            helpmap.put(i,nums[i]);
        }
        return false;
    }
}