LeetCode 220. Contains Duplicate III

原创于 2018-07-31 20:49:38 发布 · 99 阅读

CC 4.0 BY-SA版权

本文探讨了一种算法问题，即在给定数组中查找是否存在两个元素之间的差值不超过t，且下标差值不超过k。通过使用map数据结构，文章详细介绍了如何实现这一功能，并提供了完整的代码示例。

220. 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 absolute difference between nums[i] and nums[j] is at most t and the absolute difference between i and j is at most k.

Example 1:

Input: nums = [1,2,3,1], k = 3, t = 0
Output: true
Example 2:

Input: nums = [1,0,1,1], k = 1, t = 2
Output: true
Example 3:

Input: nums = [1,5,9,1,5,9], k = 2, t = 3
Output: false

Approach

题目大意：给你一组数据，问你是否存在两个数据之间差值不大于t，下标差值不大于k。（题目不难，恶心的是数据，超int，t为负数等等。。。改了代码好几次）
思路方法：首先将一些违法的数据先返回。采用map，key装数据，value装下标，然后我们要维持下标范围，然后每次用lower_bound查找不小于（nums[i]-t）的数据。

Code

class Solution {
public:
    bool containsNearbyAlmostDuplicate(vector<int>& nums, int k, int t) {
        if (nums.size() == 0)return false;
        if (t < 0)return false;
        map<int, int> mp;
        int j = 0;
        for (int i = 0; i < nums.size(); ++i) {
            if (i - j > k) {
                mp.erase(nums[j++]);
            }
            map<int,int>::iterator it = mp.lower_bound((long long)nums[i] - t);
            if (it != mp.end() && abs(it->first - nums[i]) <= t) {
                return true;
            }
            mp[nums[i]] = i;
        }
        return false;
    }
};