LC 846. Hand of Straights
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 <= hand.length <= 10000
0 <= hand[i] <= 10^9
1 <= W <= hand.length
Runtime: 80 ms, faster than 24.90% of C++ online submissions for Hand of Straights.
我的思路很直接,用map保存。
class Solution {
public:
bool isNStraightHand(vector<int>& hand, int W) {
int n = hand.size();
if (n < 1 || W < 1 || n < W)
{
return false;
}
if (n % W != 0)
{
return false;
}
map<int,int> mp;
for(auto x : hand) mp[x]++;
for(auto it : mp){
if(it.second == 0) continue;
int cntit = it.second;
for(int i=1; i<W;i++){
if(!mp.count(it.first) || mp[it.first+i] < mp[it.first]) return false;
mp[it.first+i] -= mp[it.first];
}
mp[it.first] = 0;
}
for(auto it : mp){
if(it.second != 0) return false;
}
return true;
}
};
网上看到一个比较好的思路是,把所有的数%W,因为是连续的,所以一个连续的W个数modW后必定在0 ~ W-1 中是连续存在的,妙。
class Solution {
public:
bool isNStraightHand(vector<int>& hand, int W) {
int n = hand.size();
if (n < 1 || W < 1 || n < W)
{
return false;
}
if (n % W != 0)
{
return false;
}
vector<int> count (W, 0);
for (const int& h : hand)
{
++count[h % W];
}
int expect = count.front();
for(const int& c : count)
{
if (c != expect)
{
return false;
}
}
return true;
}
};