add 846

README.md

@@ -299,6 +299,7 @@ My LeetCode solutions with Chinese explanation. 我的LeetCode中文题解。
 | 665 |[Non-decreasing Array](https://leetcode.com/problems/non-decreasing-array)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/665.%20Non-decreasing%20Array.md)|Easy|  |
 | 714 |[Best Time to Buy and Sell Stock with Transaction Fee](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/714.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock%20with%20Transaction%20Fee.md)|Medium|  |
 | 829 |[Consecutive Numbers Sum](https://leetcode.com/problems/consecutive-numbers-sum/)|[C++](solutions/829.%20Consecutive%20Numbers%20Sum.md)|Hard|  |
+| 846 |[Hand of Straights](https://leetcode.com/problems/hand-of-straights/)|[C++](solutions/846.%20Hand%20of%20Straights.md)|Medium|  |
 | 905 |[Sort Array By Parity](https://leetcode.com/problems/sort-array-by-parity/)|[C++](solutions/905.%20Sort%20Array%20By%20Parity.md)|Easy|  |
 | 946 |[Validate Stack Sequences](https://leetcode.com/problems/validate-stack-sequences/)|[C++](solutions/946.%20Validate%20Stack%20Sequences.md)|Medium|  |
 | 1155 |[Number of Dice Rolls With Target Sum](https://leetcode.com/problems/number-of-dice-rolls-with-target-sum/)|[C++](solutions/1155.%20Number%20of%20Dice%20Rolls%20With%20Target%20Sum.md)|Medium|  |

solutions/846. Hand of Straights.md

@@ -0,0 +1,36 @@
+# [846. Hand of Straights](https://leetcode.com/problems/hand-of-straights/)
+
+# 思路
+
+这道题说手里有扑克牌若干，是否能将手里的牌都以顺子的形式出完，每组顺子的牌数都为W。
+
+由于顺子是等差为1的等差数列，所以如果知道了最小元素也就知道了其他所有元素。所以我们可以使用贪婪的算法：首先取出手中最小的牌，再按照不断+1的规则组成第一组顺子；再在剩下的牌中按照上一次一样的规则组成第二组顺子；以此类推。由此可见，我们需要一个快速获得最小值和快速删除元素的数据结构，map和multiset满足我们的要求。
+
+若使用map，我们可以记录元素值到出现次数的隐射，删除操作即将出现次数减1；若使用multiset，我们每次根据迭代器删除元素即可，即`.erase(.find(val))`（不能根据元素值删除`.erase(val)`，那样会将所有值相等的元素都删了），代码中我们使用multiset。
+
+时间复杂度O(nlogn)，空间复杂度O(n)
+
+# C++
+```C++
+class Solution {
+public:
+    bool isNStraightHand(vector<int>& hand, int W) {
+        int n = hand.size(), groups = n / W;
+        if(n % W) return false;
+        
+        multiset<int>cards;
+        for(int num: hand) cards.insert(num);
+        
+        int pre;
+        while(groups--){
+            pre = *cards.begin(); 
+            cards.erase(cards.begin());
+            for(int j = 1; j < W; j++){
+                if(cards.find(++pre) == cards.end()) return false;
+                cards.erase(cards.find(pre));
+            }
+        }
+        return true;
+    }
+};
+```