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.
按照题意使用两层for循环进行判断。结果超时了。附上超时代码,便于理解题意。
class Solution { public: bool containsNearbyDuplicate(vector<int>& nums, int k) { int n = nums.size(); if (n < 2 || k <= 0) return false; for (int i = 0; i < n; i++) { for (int j = i + 1; j <= i + k && j < n; j++) { if (nums[i] == nums[j]) return true; } } return false; } }; // TLE
使用map来存储元素值与索引之间的关系,键:元素值,值:索引值。
遍历数组中的每一个元素,如果可以找到map中元素,判断map中元素与此时数组元素索引之差是否小于等于k。
class Solution { public: bool containsNearbyDuplicate(vector<int>& nums, int k) { int n = nums.size(); if (n < 2 || k <= 0) return false; unordered_map<int, int> m; for (int i = 0; i < n; i++) { if (m.count(nums[i]) != 0 && i - m[nums[i]] <= k) return true; else m[nums[i]] = i; } return false; } }; // 22 ms