[LeetCode 220] Contains Duplicate 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

分析

这一题暴力搜索的方法会超时。对于这种滑动窗口的题目，首先应该想到使用multiset进行解决。multiset用来存放窗口[i, k+i]的所有的数字，当窗口向后移动时，需要将nums[i]从multiset中删除，注意删除时，需要使用erase(s.find(nums[i]))，而不能直接使用erase(nums[i])。此时需要新插入的数字为nums[i, k+i+1]。我们要查看multiset中是否存在数字x使得 -t <= x - nums[k+i+1] <=t，那么转换一下就是nums[i+k+1] - t <= x <= nums[i+k+1] + t。这样就可以借助lower_bound函数在multiset中寻找了。

Code

class Solution { public: bool containsNearbyAlmostDuplicate(vector<int>& nums, int k, int t) { int len = nums.size(); if (len < 2) return false; multiset<long long> s; s.insert(nums[0]); for (int i = 1; i < len; i ++) { if (s.size() == k + 1) s.erase(s.find(nums[i-k-1])); long long low = (long long)nums[i] - t; long long up = (long long)nums[i] + t; multiset<long long>::iterator iter; iter = s.lower_bound(low); if (iter != s.end() && *iter <= up) return true; s.insert(nums[i]); } return false; } };

 

运行效率 

Runtime: 20 ms, faster than 81.38% of C++ online submissions for Contains Duplicate III.

Memory Usage: 10.9 MB, less than 38.76% of C++ online submissions forContains Duplicate III.