标签：ret 排序 span tween i++ lis ash content 情况
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] and nums[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) { if (t < 0 || k < 1){ return false; } if (k > nums.length - 1){ k = nums.length - 1; } ArrayList<Long> list = new ArrayList(); for (int i = 0; i <= k; i++){ list.add((long)nums[i]); } Collections.sort(list); for (int i = 1; i < list.size(); i++){ if (list.get(i) - list.get(i - 1) <= t){ return true; } } for (int i = k + 1; i < nums.length; i++){ list.remove((long)nums[i - k - 1]); list.add((long)nums[i]); Collections.sort(list); for (int j = 1; j < list.size(); j++){ if (list.get(j) - list.get(j - 1) <= t){ return true; } } } return false; } }
2、用map实现，类似于桶排序的原理，首先把数据都转成正数，并且使用long型（否则会出错，比如-2和2除以3都等于0，但是相距4大于3），同时桶（t） = t + 1（t == 0的情况要排出）
public class Solution { public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) { if (k < 1 || t < 0){ return false; } HashMap<Long, Long> map = new HashMap(); for (int i = 0; i < nums.length; i++){ long pos = (long) nums[i] - Integer.MIN_VALUE; long num = pos / ((long) t + 1); if (map.containsKey(num) || (map.containsKey(num - 1) && Math.abs(map.get(num - 1) - pos) <= t) || (map.containsKey(num + 1) && Math.abs(map.get(num + 1) - pos) <= t)){ return true; } if (map.size() >= k){ long delete = ((long) nums[i - k] - Integer.MIN_VALUE) / ((long) t + 1); map.remove(delete); } map.put(num, pos); } return false; } }
leetcode 220. Contains Duplicate III 求一个数组中有没有要求的元素 ---------- java
标签：ret 排序 span tween i++ lis ash content 情况
原文地址：http://www.cnblogs.com/xiaoba1203/p/6739411.html
