[LeetCode] 220. Contains Duplicate III 包含重复元素 III

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

给定一个数组和两个整数t 和k ，求是否有不同的两个下标i和j，满足|nums[i] – nums[j]|<= t && | i – j | <=k

解法：参考： 细语呢喃

Java:

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

Python:

class Solution:
    # @param {integer[]} nums
    # @param {integer} k
    # @param {integer} t
    # @return {boolean}
    def containsNearbyAlmostDuplicate(self, nums, k, t):
        if k < 0 or t < 0:
            return False
        window = collections.OrderedDict()
        for n in nums:
            # Make sure window size
            if len(window) > k:
                window.popitem(False)
                
            bucket = n if not t else n // t
            # At most 2t items.
            for m in (window.get(bucket - 1), window.get(bucket), window.get(bucket + 1)):
                if m is not None and abs(n - m) <= t:
                    return True
            window[bucket] = n
        return False　

C++:

class Solution {
public:
    bool containsNearbyAlmostDuplicate(vector<int>& nums, int k, int t) {
        map<long long, int> m;
        int j = 0;
        for (int i = 0; i < nums.size(); ++i) {
            if (i - j > k) m.erase(nums[j++]);
            auto a = m.lower_bound((long long)nums[i] - t);
            if (a != m.end() && abs(a->first - nums[i]) <= t) return true;
            m[nums[i]] = i;
        }
        return false;
    }
};

　　

类似题目：

[LeetCode] 217. Contains Duplicate 包含重复元素

[LeetCode] 219. Contains Duplicate II 包含重复元素 II

　　

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