Hard题

问题描述

Given an unsorted array of integers, find the length of the longest consecutive elements sequence.
For example,
Given [100, 4, 200, 1, 3, 2],
The longest consecutive elements sequence is [1, 2, 3, 4]. Return its length: 4.
Your algorithm should run in O(n) complexity.

问题分析

其实想到了类似的思路，但是没有打通。
构建一个map，用这个map保存包含当前值的最长连续串的长度。每次新读入一个数据x时，要看x-1和x+1是否存在，如果存在，则加上两边的长度。具体看代码中的注释吧~

AC代码

class Solution {
public:
    int longestConsecutive(vector<int>& nums) {
        unordered_map<int, int> m;
        int max = 0;
        for (int i = 0; i < nums.size(); i++){
            int num = nums[i];
            if (m.find(num) == m.end()){
                int left = 0;
                int right = 0;
                if (m.find(num-1) != m.end())
                    left = m[num-1];
                if (m.find(num+1) != m.end())
                    right = m[num+1];
                //下面三行是关键。不用将当前串中所有最长值都改变，只要将边界改变即可！
                m[num] = left + 1 + right;
                m[num-left] = m[num];
                m[num+right] = m[num];
                if (m[num] > max)
                    max = m[num];
            }
        }
        return max;
    }
};