feat: add solutions to lc problems: No.1347.Minimum Number of Steps to Make Two Strings Anagram

yanglbme · yanglbme · commit bb19e78fabfe · 2021-09-28T20:24:44.000+08:00
diff --git a/solution/1300-1399/1347.Minimum Number of Steps to Make Two Strings Anagram/README.md b/solution/1300-1399/1347.Minimum Number of Steps to Make Two Strings Anagram/README.md
@@ -57,27 +57,92 @@
 	<li><code>s</code> 和 <code>t</code>&nbsp;只包含小写英文字母</li>
 </ul>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->
 
+“哈希表”实现。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def minSteps(self, s: str, t: str) -> int:
+        counter = collections.Counter(s)
+        res = 0
+        for c in t:
+            if counter[c] > 0:
+                counter[c] -= 1
+            else:
+                res += 1
+        return res
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    public int minSteps(String s, String t) {
+        int[] counter = new int[26];
+        for (char c : s.toCharArray()) {
+            ++counter[c - 'a'];
+        }
+        int res = 0;
+        for (char c : t.toCharArray()) {
+            if (counter[c - 'a'] > 0) {
+                --counter[c - 'a'];
+            } else {
+                ++res;
+            }
+        }
+        return res;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int minSteps(string s, string t) {
+        vector<int> counter(26);
+        for (char c : s) ++counter[c - 'a'];
+        int res = 0;
+        for (char c : t)
+        {
+            if (counter[c - 'a'] > 0) --counter[c - 'a'];
+            else ++res;
+        }
+        return res;
+    }
+};
+```
 
+### **Go**
+
+```go
+func minSteps(s string, t string) int {
+	counter := make([]int, 26)
+	for _, c := range s {
+		counter[c-'a']++
+	}
+	res := 0
+	for _, c := range t {
+		if counter[c-'a'] > 0 {
+			counter[c-'a']--
+		} else {
+			res++
+		}
+	}
+	return res
+}
 ```
 
 ### **...**
diff --git a/solution/1300-1399/1347.Minimum Number of Steps to Make Two Strings Anagram/README_EN.md b/solution/1300-1399/1347.Minimum Number of Steps to Make Two Strings Anagram/README_EN.md
@@ -58,21 +58,84 @@
 	<li><code>s</code> and <code>t</code> contain lower-case English letters only.</li>
 </ul>
 
-
 ## Solutions
 
 <!-- tabs:start -->
 
 ### **Python3**
 
 ```python
-
+class Solution:
+    def minSteps(self, s: str, t: str) -> int:
+        counter = collections.Counter(s)
+        res = 0
+        for c in t:
+            if counter[c] > 0:
+                counter[c] -= 1
+            else:
+                res += 1
+        return res
 ```
 
 ### **Java**
 
 ```java
+class Solution {
+    public int minSteps(String s, String t) {
+        int[] counter = new int[26];
+        for (char c : s.toCharArray()) {
+            ++counter[c - 'a'];
+        }
+        int res = 0;
+        for (char c : t.toCharArray()) {
+            if (counter[c - 'a'] > 0) {
+                --counter[c - 'a'];
+            } else {
+                ++res;
+            }
+        }
+        return res;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int minSteps(string s, string t) {
+        vector<int> counter(26);
+        for (char c : s) ++counter[c - 'a'];
+        int res = 0;
+        for (char c : t)
+        {
+            if (counter[c - 'a'] > 0) --counter[c - 'a'];
+            else ++res;
+        }
+        return res;
+    }
+};
+```
 
+### **Go**
+
+```go
+func minSteps(s string, t string) int {
+	counter := make([]int, 26)
+	for _, c := range s {
+		counter[c-'a']++
+	}
+	res := 0
+	for _, c := range t {
+		if counter[c-'a'] > 0 {
+			counter[c-'a']--
+		} else {
+			res++
+		}
+	}
+	return res
+}
 ```
 
 ### **...**
diff --git a/solution/1300-1399/1347.Minimum Number of Steps to Make Two Strings Anagram/Solution.cpp b/solution/1300-1399/1347.Minimum Number of Steps to Make Two Strings Anagram/Solution.cpp
@@ -0,0 +1,14 @@
+class Solution {
+public:
+    int minSteps(string s, string t) {
+        vector<int> counter(26);
+        for (char c : s) ++counter[c - 'a'];
+        int res = 0;
+        for (char c : t)
+        {
+            if (counter[c - 'a'] > 0) --counter[c - 'a'];
+            else ++res;
+        }
+        return res;
+    }
+};
diff --git a/solution/1300-1399/1347.Minimum Number of Steps to Make Two Strings Anagram/Solution.go b/solution/1300-1399/1347.Minimum Number of Steps to Make Two Strings Anagram/Solution.go
@@ -0,0 +1,15 @@
+func minSteps(s string, t string) int {
+	counter := make([]int, 26)
+	for _, c := range s {
+		counter[c-'a']++
+	}
+	res := 0
+	for _, c := range t {
+		if counter[c-'a'] > 0 {
+			counter[c-'a']--
+		} else {
+			res++
+		}
+	}
+	return res
+}
diff --git a/solution/1300-1399/1347.Minimum Number of Steps to Make Two Strings Anagram/Solution.java b/solution/1300-1399/1347.Minimum Number of Steps to Make Two Strings Anagram/Solution.java
@@ -1,18 +1,16 @@
-import java.util.*;
-
 class Solution {
     public int minSteps(String s, String t) {
-        Map<Character, Integer> map = new HashMap<>();
-        for (char c : t.toCharArray()) {
-            if (map.containsKey(c)) {
-                map.put(c, map.get(c) + 1);
-            } else map.put(c, 1);
+        int[] counter = new int[26];
+        for (char c : s.toCharArray()) {
+            ++counter[c - 'a'];
         }
         int res = 0;
-        for (char c : s.toCharArray()) {
-            if (map.containsKey(c) && map.get(c) > 0) {
-                map.put(c, map.get(c) - 1);
-            } else res ++;
+        for (char c : t.toCharArray()) {
+            if (counter[c - 'a'] > 0) {
+                --counter[c - 'a'];
+            } else {
+                ++res;
+            }
         }
         return res;
     }
diff --git a/solution/1300-1399/1347.Minimum Number of Steps to Make Two Strings Anagram/Solution.py b/solution/1300-1399/1347.Minimum Number of Steps to Make Two Strings Anagram/Solution.py
@@ -0,0 +1,10 @@
+class Solution:
+    def minSteps(self, s: str, t: str) -> int:
+        counter = collections.Counter(s)
+        res = 0
+        for c in t:
+            if counter[c] > 0:
+                counter[c] -= 1
+            else:
+                res += 1
+        return res
