Contains Duplicate III
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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 difference between nums[i] and nums[j] is at most t and the difference between i and j is at most k.
一开始理解错题目了,还以为是要求所有的数对都必须符合要求,一直超时。其实题目是问是不是存在一个符合条件的数对,所以就容易多了,维护一个大小为 k 的二叉搜索树,来一个新的元素时,在BST上二分搜索有没有符合条件的数对,动态更新这个BST。因为BST的大小为 k 或不超过 k,所以这里面的数下标的差值一定是符合条件的。还有几点要注意的就是nums[i]与nums[j]的差值的是绝对值,所以要分别找lower_bound跟upper_bound,数据比较坑爹,为了防止溢出,容器用long long类型的。
1 class Solution { 2 public: 3 bool containsNearbyAlmostDuplicate(vector<int>& nums, int k, int t) { 4 multiset<long long> bst; 5 for (int i = 0; i < nums.size(); ++i) { 6 if (bst.size() == k + 1) bst.erase(bst.find(nums[i - k - 1])); 7 auto lb = bst.lower_bound(nums[i]); 8 if (lb != bst.end() && abs(*lb - nums[i]) <= t) return true; 9 auto ub = bst.upper_bound(nums[i]);10 if (ub != bst.begin() && abs(*(--ub) - nums[i]) <= t) return true;11 bst.insert(nums[i]);12 }13 return false;14 }15 };
不用比较两次,直接找nums[i] - t的lower_bound, 这个值就是与nums[i]差值最近的值。
1 class Solution { 2 public: 3 bool containsNearbyAlmostDuplicate(vector<int>& nums, int k, int t) { 4 multiset<long long> bst; 5 for (int i = 0; i < nums.size(); ++i) { 6 if (bst.size() == k + 1) bst.erase(bst.find(nums[i - k - 1])); 7 auto lb = bst.lower_bound(nums[i] - t); 8 if (lb != bst.end() && *lb - nums[i] <= t) return true; 9 bst.insert(nums[i]);10 }11 return false;12 }13 };
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