LeetCode 219.Contains Duplicate II
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219.Contains Duplicate II
一、问题描述
Given an array of integers and an integer k, find out whether there are two distinct indices i and j in the array such that nums[i] = nums[j]*and the **absolute difference between *i and j is at most k.
二、输入输出
三、解题思路
- 使用
map
unordered_map
set
unordered_set
multi_map
可以做这道题 - 思路1:遍历vector 将
<value,index>
存储在map中,每次添加之前判断map是否已经存在,如果存在再判断两个index是否相差<=k 如果是,返回true,如果不是更新map中该key的value为最新值
bool containsNearbyDuplicate(vector<int>& nums, int k) { unordered_map<int, int> dict; int n = nums.size(); for (int i = 0; i < n; ++i) { if( dict.find(nums[i]) == dict.end() ) dict.insert(make_pair(nums[i], i)); else{ int index = dict.find(nums[i])->second; if(abs(index - i) <= k) return true; else dict[nums[i]] = i; } } return false; }
- 思路2:使用set或者map 但是里面不是保存所有的,里面只保存k个元素,一旦
i>k
就把开头超过的元素删除。假设当前遍历到index,并且添加到set中了,那么set中保存的就是[index - k, index - 1]
一共k个元素,所以这时候该删除的元素是index - k - 1
bool containsNearbyDuplicate(vector<int>& nums, int k) { if (k <= 0) return false; if (k > nums.size()) k = nums.size(); unordered_set<int> s; for (int i = 0; i < nums.size(); ++i) { if (i > k) s.erase(nums[i - k - 1]); if (s.find(nums[i]) != s.end()) return true; s.insert(nums[i]); } return false; }
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