Contains Duplicate III (M)
题目
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.
Example 1:
Input: nums = [1,2,3,1], k = 3, t = 0
Output: true
Example 2:
Input: nums = [1,0,1,1], k = 1, t = 2
Output: true
Example 3:
Input: nums = [1,5,9,1,5,9], k = 2, t = 3
Output: false
题意
判断数组中是否存在这样一组数，它们的值之差的绝对值不大于t，且下标之差的绝对值不大于k。
思路
最简单的方法是直接遍历，再判断当前数与它之后的k个数的差值是否符合条件。
优化的方法是借助红黑树实现滑动窗口思想：在红黑树中只保存k+1个连续的数(这样就控制了树中所有数的下标差不大于k)，每次加入新数时，先去除一个数(容量已到达k+1的情况下)，在树中找到小于新数的最大数和大于新数的最小数，判断它们与新数的差是否满足条件，最后将新数加入红黑树中。
代码实现
Java
暴力法
class Solution {
public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
if (t < 0) {
return false;
}
for (int i = 0; i < nums.length; i++) {
for (int j = i + 1; j <= Math.min(i + k, nums.length - 1); j++) {
// 不转化为long可能会有溢出
if (Math.abs((long) nums[i] - (long) nums[j]) <= (long) t) {
return true;
}
}
}
return false;
}
}
红黑树
class Solution {
public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
if (t < 0) {
return false;
}
TreeSet<Integer> set = new TreeSet<>();
for (int i = 0; i < nums.length; i++) {
// 保证树的容量始终不超过k+1
if (i > k) {
set.remove(nums[i - k - 1]);
}
if (set.contains(nums[i])) {
return true;
}
// 因为可能为null，所以必须用包装类接收
Integer lower = set.lower(nums[i]);
Integer higher = set.higher(nums[i]);
// 转换成long计算防止溢出
if (lower != null && (long) nums[i] - lower <= t
|| higher != null && (long) higher - nums[i] <= t) {
return true;
}
set.add(nums[i]);
}
return false;
}
}
JavaScript
暴力法
/**
* @param {number[]} nums
* @param {number} k
* @param {number} t
* @return {boolean}
*/
var containsNearbyAlmostDuplicate = function (nums, k, t) {
for (let i = 0; i < nums.length; i++) {
for (let j = i + 1; j <= i + k && j < nums.length; j++) {
if (Math.abs(nums[j] - nums[i]) <= t) {
return true
}
}
}
return false
}