49 Anagrams

Author: int32bit

README.md

@@ -32,6 +32,7 @@
 + [43 Multiply Strings(大数乘法)](algorithms/MultiplyStrings)
 + [46 Permutations(全排列)](algorithms/Permutations)
 + [47 Permutations II(全排列)](algorithms/Permutations2)
++ [49 Anagrams(字符串，回文构词)](algorithms/Anagrams)
 + [50 Pow(x, n)(二分，浮点数比较，INT\_MIN绝对值)](algorithms/Pow)
 + [58 Length of Last Word](algorithms/LengthofLastWord)
 + [60 Permutation Sequence(康托展开)](algorithms/PermutationSequence)

algorithms/Anagrams/README.md

@@ -0,0 +1,46 @@
+## Anagrams 
+
+Given an array of strings, return all groups of strings that are anagrams.
+
+Note: All inputs will be in lower-case.
+
+## Solution
+
+两个单词的字符相同，只是排列顺序不同，称为回文构词法(anagrams), 比如tea eta ate aet都是anagrams。
+
+题目要求找出所有的anagrams组，比如`["dog","cat","god","tac", "hello", "world"]`, 应该返回`["dog","cat","god","tac"]`,
+因为dog 和 god是一组， cat和tac是一组，即
+
+```
+[
+  ["dog", "god"],
+  ["cat", "tac"]
+]
+```
+
+判断两个字符串是不是anagrams，最简单的方法是对字符排序然后判断是否相等。比如`tac`、`cat`按字符排序后分别为`act`、`act`，显然是相等的，因此互为anagrams.
+
+给定一个字符串数组，如何找出所有的anagrams组。当一个字符串处理时，我们需要判断之前是否出现过互为anagrams的字符串。
+如果没有出现过，我们保存到一个map中，key保存字符串的排序后的字符串，value保存第一次出现的索引。若已经出现过，并且当前是
+第二次出现，则前一次出现的和当前的字符串互为anagrams
+
+```cpp
+vector<string> anagrams(const vector<string>& strs) {
+	unordered_map<string, int> m;
+	vector<string> result;
+	for (int i = 0; i < strs.size(); ++i) {
+		string s = strs[i];
+		sort(begin(s), end(s));
+		if (m.find(s) == m.end()) {
+			m[s] = i;
+		} else {
+			if (m[s] >= 0) {
+				result.push_back(strs[m[s]]);
+				m[s] = -1;
+			}
+			result.push_back(strs[i]);
+		}
+	}
+	return result;
+}
+```

algorithms/Anagrams/solve.cpp

@@ -0,0 +1,35 @@
+#include <string>
+#include <algorithm>
+#include <iostream>
+#include <unordered_map>
+using namespace std;
+class Solution {
+	public:
+		vector<string> anagrams(const vector<string>& strs) {
+			unordered_map<string, int> m;
+			vector<string> result;
+			for (int i = 0; i < strs.size(); ++i) {
+				string s = strs[i];
+				sort(begin(s), end(s));
+				if (m.find(s) == m.end()) {
+					m[s] = i;
+				} else {
+					if (m[s] >= 0) {
+						result.push_back(strs[m[s]]);
+						m[s] = -1;
+					}
+					result.push_back(strs[i]);
+				}
+			}
+			return result;
+		}
+};
+int main(int argc, char **argv)
+{
+	Solution solution;
+	vector<string> strs = {"tea", "and", "ate", "eat", "den"};
+	auto result = solution.anagrams(strs);
+	for_each(begin(result), end(result), [](string s) {cout << s << " ";});
+	cout << endl;
+	return 0;
+}