leetcode 846. Hand of Straights

leetcode 846. Hand of Straights

题目：

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

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解法：

这个题大致意思是让我们判断能否在给定序列中找到 hand.size() / W个长度为 W 的连续递增序列。

• 首先，我们可以判断 hand.size() % W是否成立，因为任意一个不满足 W 个的序列都不满足题意

• 其次，我们就要考虑怎么凑成连续的序列：

  • 这里我们可以考虑使用map映射的方法，举个例子：

    • 假如现在序列为example 1 ,也即hand = [1,2,3,6,2,3,4,7,8],那么

      • map[1] = 1;
      • map[2] = 2;
      • map[3] = 2;
      • map[4] = 1;
      • map[6] = 1;
      • map[7] = 1;
      • map[8] = 1;

    • 我们现在可以这么思考，如果序列连续，那么任意一个开始点 start 往后数W-1个连续数字的map值一定满足：
      $map[start + W - 1] >= map[start + W - 2] >= ...... map[start]$

    • 且每一次我们处理完相应的序列之后，都要让的map[start]之前的map[start + W - N]的值减去map[start]，这表明我们已经访问过某个序列。如果在处理过程中有某个map[start + W - N] < 0，那证明序列出现了中断，也就不满足题意。

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代码：

class Solution {
public:
    bool isNStraightHand(vector<int>& hand, int W) {
        int len = hand.size();
        if(len%W != 0)
            return false;
        sort(hand.begin(), hand.end());
        map<int, int> cnt;
        for(auto h:hand)
            cnt[h]++;
        for(auto it:cnt){
            if(it.second > 0){
                for(int i=W-1; i>=0; i--){
                    cnt[it.first+i] -= cnt[it.first];
                    if(cnt[it.first+i] < 0)
                        return false;
                }
            }
        }
        return true;
    }
};