标签：cto 简单 duplicate problem summary 循环 i++ 存在 hashmap
Problem:
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 difference between i and j is at most k.
Analysis:
给出数组nums，整型数k，找到数组中是否存在nums[i]和nums[j]相等且i - j小于k的情况。
Summary:
开始用最简单写法，两重for循环逐一查找，出现TLE。
后用HashMap改进，key为数组中出现的数字，value为最近出现位置的下标。
1 class Solution { 2 public: 3 bool containsNearbyDuplicate(vector<int>& nums, int k) { 4 int len = nums.size(); 5 unordered_map<int, int> m; 6 7 for (int i = 0; i < len; i++) { 8 if (m.find(nums[i]) != m.end() && i - m[nums[i]] <= k) { 9 return true; 10 } 11 else { 12 m[nums[i]] = i; 13 } 14 } 15 16 return false; 17 } 18 };
LeetCode 219 Contains Duplicate II
标签：cto 简单 duplicate problem summary 循环 i++ 存在 hashmap
原文地址：http://www.cnblogs.com/VickyWang/p/6232277.html
