220. 存在重复元素 III

English Version

题目描述

给你一个整数数组 nums 和两个整数 k 和 t 。请你判断是否存在 两个不同下标 i 和 j，使得 abs(nums[i] - nums[j]) <= t ，同时又满足 abs(i - j) <= k 。

如果存在则返回 true，不存在返回 false。

 

示例 1：

输入：nums = [1,2,3,1], k = 3, t = 0
输出：true

示例 2：

输入：nums = [1,0,1,1], k = 1, t = 2
输出：true

示例 3：

输入：nums = [1,5,9,1,5,9], k = 2, t = 3
输出：false

 

提示：

• 0 <= nums.length <= 2 * 104
• -231 <= nums[i] <= 231 - 1
• 0 <= k <= 104
• 0 <= t <= 231 - 1

解法

“滑动窗口 + 有序集合”实现。

Python3

from sortedcontainers import SortedSet


class Solution:
    def containsNearbyAlmostDuplicate(self, nums: List[int], k: int, t: int) -> bool:
        s = SortedSet()
        for i, num in enumerate(nums):
            idx = s.bisect_left(num - t)
            if 0 <= idx < len(s) and s[idx] <= num + t:
                return True
            s.add(num)
            if i >= k:
                s.remove(nums[i - k])
        return False

Java

class Solution {
    public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
        TreeSet<Long> ts = new TreeSet<>();
        for (int i = 0; i < nums.length; ++i) {
            Long x = ts.ceiling((long) nums[i] - (long) t);
            if (x != null && x <= (long) nums[i] + (long) t) {
                return true;
            }
            ts.add((long) nums[i]);
            if (i >= k) {
                ts.remove((long) nums[i - k]);
            }
        }
        return false;
    }
}

C++

class Solution {
public:
    bool containsNearbyAlmostDuplicate(vector<int>& nums, int k, int t) {
        set<long> s;
        for (int i = 0; i < nums.size(); ++i)
        {
            auto it = s.lower_bound((long) nums[i] - t);
            if (it != s.end() && *it <= (long) nums[i] + t) return true;
            s.insert((long) nums[i]);
            if (i >= k) s.erase((long) nums[i - k]);
        }
        return false;
    }
};

Go

func containsNearbyAlmostDuplicate(nums []int, k int, t int) bool {
	n := len(nums)
	left, right := 0, 0
	rbt := redblacktree.NewWithIntComparator()
	for right < n {
		cur := nums[right]
		right++
		if p, ok := rbt.Floor(cur); ok && cur-p.Key.(int) <= t {
			return true
		}
		if p, ok := rbt.Ceiling(cur); ok && p.Key.(int)-cur <= t {
			return true
		}
		rbt.Put(cur, struct{}{})
		if right-left > k {
			rbt.Remove(nums[left])
			left++
		}
	}
	return false
}

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