219. Contains Duplicate II

原创于 2017-10-29 13:16:13 发布 · 202 阅读

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本文介绍了一个算法问题：如何在一个整数数组中找到是否存在两个不同的索引 i 和 j，使得 nums[i] = nums[j] 并且 i 和 j 的绝对差不超过 k。提供了 C++ 实现代码，展示了通过双重循环来解决此问题的方法。

Problems:

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.

Solutions:

#include<iostream>
#include<string>
#include<vector>

using namespace std;

class Solution {
public:
bool containsNearbyDuplicate(vector<int>& nums, int k) {
for (int i = 0; i < nums.size(); i++)
{
for (int j = i + 1; j < nums.size(); j++)
{
if (j - i > k)
break;
if (nums[i] == nums[j])
return true;
}
}
return false;
}
};

int main()
{
Solution s;
vector<int> nums;
int k;
cin >> k;
nums.push_back(1);
nums.push_back(2);
nums.push_back(3);
nums.push_back(4);
nums.push_back(5);
nums.push_back(1);
bool b;
b = s.containsNearbyDuplicate(nums, k);
cout << s.containsNearbyDuplicate(nums, k) << endl;
int n;
cin >> n;
return 0;
}