Merge pull request #1670 from chenzhg-maker/master

Update 0583.两个字符串的删除操作 C++

problems/0072.编辑距离.md

@@ -362,6 +362,33 @@ function minDistance(word1: string, word2: string): number {
 };
 ```
 
+C：
+
+```c
+int min(int num1, int num2, int num3) {
+    return num1 > num2 ? (num2 > num3 ? num3 : num2) : (num1 > num3 ? num3 : num1); 
+}
+
+int minDistance(char * word1, char * word2){
+    int dp[strlen(word1)+1][strlen(word2)+1];
+    dp[0][0] = 0;
+    for (int i = 1; i <= strlen(word1); i++) dp[i][0] = i;
+    for (int i = 1; i <= strlen(word2); i++) dp[0][i] = i;
+
+    for (int i = 1; i <= strlen(word1); i++) {
+        for (int j = 1; j <=strlen(word2); j++) {
+            if (word1[i-1] == word2[j-1]) {
+                dp[i][j] = dp[i-1][j-1];
+            }
+            else {
+                dp[i][j] = min(dp[i-1][j-1], dp[i][j-1], dp[i-1][j]) + 1;
+            }
+        }
+    }
+    return dp[strlen(word1)][strlen(word2)];
+}
+```
+
 
 
 -----------------------

problems/0583.两个字符串的删除操作.md

@@ -47,6 +47,8 @@ dp[i][j]：以i-1为结尾的字符串word1，和以j-1位结尾的字符串word
 
 那最后当然是取最小值，所以当word1[i - 1] 与 word2[j - 1]不相同的时候，递推公式：dp[i][j] = min({dp[i - 1][j - 1] + 2, dp[i - 1][j] + 1, dp[i][j - 1] + 1});
 
+因为dp[i - 1][j - 1] + 1等于 dp[i - 1][j] 或 dp[i][j - 1]，所以递推公式可简化为：dp[i][j] = min(dp[i - 1][j] + 1, dp[i][j - 1] + 1);
+
 
 3. dp数组如何初始化
 
@@ -90,7 +92,7 @@ public:
                 if (word1[i - 1] == word2[j - 1]) {
                     dp[i][j] = dp[i - 1][j - 1];
                 } else {
-                    dp[i][j] = min({dp[i - 1][j - 1] + 2, dp[i - 1][j] + 1, dp[i][j - 1] + 1});
+                    dp[i][j] = min(dp[i - 1][j] + 1, dp[i][j - 1] + 1);
                 }
             }
         }