LeetCode #846 - Hand of Straights

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本文介绍了一种手牌分组算法，旨在将一组整数卡片重新排列为大小相等且连续的子组。通过使用map记录每个数出现的次数，并从最小的数字开始删除，判断是否可以将手牌完美分组。

题目描述：

Alice has a hand of cards, given as an array of integers.

Now she wants to rearrange the cards into groups so that each group is size W, and consists of W consecutive cards.

Return true if and only if she can.

Example 1:

Input: hand = [1,2,3,6,2,3,4,7,8], W = 3

Output: true

Explanation: Alice's hand can be rearranged as [1,2,3],[2,3,4],[6,7,8].

Example 2:

Input: hand = [1,2,3,4,5], W = 4

Output: false

Explanation: Alice's hand can't be rearranged into groups of 4.

Note:

1. 1 <= hand.length <= 10000

2. 0 <= hand[i] <= 10^9

3. 1 <= W <= hand.length

用map记录每个数出现的次数，采用map的原因是map按照key从小到大排序，那么map.begin()对应最小的数字，我们就从最小的数字开始从map中删除，如果能全部删除说明可行。

class Solution {
public:
    bool isNStraightHand(vector<int>& hand, int W) {
        int n=hand.size();
        if(n%W!=0) return false;
        map<int,int> m;
        for(int i=0;i<hand.size();i++)
        {
            if(m.count(hand[i])==0) m[hand[i]]=1;
            else m[hand[i]]++;
        }
        while(!m.empty())
        {
            int x=m.begin()->first;
            for(int i=x;i<x+W;i++)
            {
                if(m.count(i)==0) return false;
                else
                {
                    m[i]--;
                    if(m[i]==0) m.erase(i);
                }
            }
        }
        return true;
    }
};