solution_659

LeetCode problem 659 description

You are given an integer array sorted in ascending order (may contain duplicates), you need to split them into several subsequences, where each subsequences consist of at least 3 consecutive integers. Return whether you can make such a split.

Example 1:

Input: [1,2,3,3,4,5]Output: TrueExplanation:You can split them into two consecutive subsequences : 1, 2, 33, 4, 5

Example 2:

Input: [1,2,3,3,4,4,5,5]Output: TrueExplanation:You can split them into two consecutive subsequences : 1, 2, 3, 4, 53, 4, 5

Example 3:

Input: [1,2,3,4,4,5]Output: False

Note:

1. The length of the input is in range of [1, 10000]

Solution

I just use a simple method to work it out. At first I record the frequency that each number appearing. Then I record the frequency of the last number of the consecutive subsequence. Here is the code:

#include <iostream>#include <vector>#include <unordered_map>using namespace std;class Solution {public:    bool isPossible (vector<int> & nums);};bool Solution::isPossible(vector<int> & nums) {    unordered_map<int, int> frequency, tails;    //Caculating the frequency of the numbers    for (int i : nums) {        frequency[i]++;    }    for (int i : nums) {        if (frequency[i] == 0) {            continue;        }        frequency[i]--;        //This means that number i is the tail of one subsequence with number        //i - 1 as the tail.        if (tails[i - 1] > 0) {            tails[i - 1]--;            tails[i]++;        }        //This means that number i is the first number of the consecutive         //subsequence with length not less than 3        else if (frequency[i + 1] > 0 && frequency[i + 2] > 0){            tails[i + 2]++;            frequency[i + 1]--;            frequency[i + 2]--;        }        else {            return false;        }    }    return true;}