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Author: assassint2017

Edit Distance/Edit_Distance.cpp

@@ -0,0 +1,119 @@
+// 思路一 自顶向下的备忘录动态规划
+// 如果不使用备忘录就会超时
+// 16ms 8.41%
+
+#include<string>
+using namespace std;
+
+class Solution {
+public:
+    int minDistance(string word1, string word2) 
+    {
+        // 申请空间，这里注意一下二维动态数组的申请
+        int **steps = new int*[word1.length() + 1];
+        for (int i = 0; i <= word1.length(); i++)
+            steps[i] = new int[word2.length() + 1];
+        
+        // 初始化
+        for (int i = 0; i <= word1.length(); i++)
+            for (int j = 0; j <= word2.length(); j++)
+                steps[i][j] = -1;
+        
+        // 动态规划
+        int res = dp(word1, word2, steps);
+        
+        // 释放空间
+        delete[] steps;
+
+        return res;
+    }
+
+    int dp(string word1, string word2, int **steps)
+    {
+        if (steps[word1.length()][word2.length()] == - 1)
+        {
+            // 判断两个字符串是否为空
+            if (word1 == "" && word2 == "")
+                steps[word1.length()][word2.length()] = 0;
+            else if (word1 == "")
+                steps[word1.length()][word2.length()] = word2.length();
+            else if (word2 == "")
+                steps[word1.length()][word2.length()] = word1.length();
+            else
+            {
+                // 如果两者都不为空的情况下进行递归的判断
+                // 如果两个字符串的最后一位是相等的
+                if (word2[word2.length() - 1] == word1[word1.length() - 1])
+                    steps[word1.length()][word2.length()] = dp(word1.substr(0, word1.length() - 1), word2.substr(0, word2.length() - 1), steps);
+                
+                // 不相等
+                else
+                {
+                    // 删除最后一个
+                    int stepDelete = dp(word1.substr(0, word1.length() - 1), word2.substr(0, word2.length()), steps);
+
+                    // 替换最后一个
+                    int stepReplace = dp(word1.substr(0, word1.length() - 1), word2.substr(0, word2.length() - 1), steps);
+
+                    // 在最后一位上插入
+                    int stepInsert = dp(word1.substr(0, word1.length()), word2.substr(0, word2.length() - 1), steps);
+
+                    steps[word1.length()][word2.length()] = minThree(stepDelete, stepReplace, stepInsert) + 1;
+                }
+            }
+        }
+        return steps[word1.length()][word2.length()];
+    }
+
+    inline int minThree(int a, int b, int c)
+    {
+        int temp = a < b ? a : b;
+        return temp < c ? temp : c;
+    }
+};
+
+
+// 思路二 自底向上的递推动规 节省空间版本
+// 4ms 100%
+
+class Solution {
+public:
+    int minDistance(string word1, string word2) 
+    {
+        // 初始化第一行的结果
+        int *temp = new int[word2.length() + 1];
+        int *steps = new int[word2.length() + 1];
+        for (int i = 0; i <= word2.length(); i++)
+            temp[i] = i;
+
+        for (int i = 1; i <= word1.length(); i++)
+        {
+            steps[0] = i;
+            for (int j = 1; j <= word2.length(); j++)
+            {
+                if (word1[i - 1] == word2[j - 1])
+                    steps[j] = temp[j - 1];
+                else
+                {
+                    int stepDelete = temp[j];
+                    int stepReplace = temp[j - 1];
+                    int stepInsert = steps[j - 1];
+
+                    int temp = stepDelete < stepReplace ? stepDelete : stepReplace;
+                    steps[j] = 1 + (temp < stepInsert ? temp : stepInsert);
+                }
+            }
+
+            for (int j = 0; j <= word2.length(); j++)
+                temp[j] = steps[j];
+        }
+        
+        int res = temp[wpord2.length()];
+
+        // 释放空间
+        delete[] steps;
+        delete[] temp;
+
+        return res;
+    }
+};

Edit Distance/Edit_Distance.py

@@ -0,0 +1,20 @@
+# 思路二的py代码
+# 172ms 91.22%
+
+class Solution:
+    def minDistance(self, word1, word2):
+        temp = [i for i in range(len(word2) + 1)]
+        steps = [0 for _ in range(len(word2) + 1)]
+        for i in range(1, len(word1) + 1):
+            steps[0] = i
+            for j in range(1, len(word2) + 1):
+                if word1[i - 1] == word2[j - 1]:
+                    steps[j] = temp[j - 1]
+                else:
+                    stepDelete = temp[j]
+                    stepReplace = temp[j - 1]
+                    stepInsert = steps[j - 1]
+                    steps[j] = min(stepDelete, stepInsert, stepReplace) + 1
+            for j in range(len(word2) + 1):
+                temp[j] = steps[j]
+        return temp[len(word2)]