Edit Distance

Author: int32bit

README.md

@@ -25,6 +25,7 @@
 + [58 Length of Last Word](algorithms/LengthofLastWord)
 + [60 Permutation Sequence(康托展开)](algorithms/PermutationSequence)
 + [65 Valid Number(细节题,状态机)](algorithms/ValidNumber)
++ [72 Edit Distance(DP)](algorithms/EditDistance)
 + [75 Sort Colors](algorithms/SortColors)
 + [100 Same Tree](algorithms/SameTree)
 + [101 Symmetric Tree](algorithms/SymmetricTree)

algorithms/EditDistance/README.md

@@ -0,0 +1,52 @@
+## Edit Distance
+
+Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)
+
+You have the following 3 operations permitted on a word:
+```
+	a) Insert a character
+	b) Delete a character
+	c) Replace a character
+```
+
+## Solution
+
+首先word1变成word2和word2变成word1的步数一样的.
+
+假设需要把s[1..i]变成t[1..j],例如将beauty变成batyu, 所需要的最少的操作数.我们可能有3种可能:
+
+1. 把s[1..i] 转化成t[1..j-1], 设操作次数为case1,我们只需要把t[j]加到后面即可，因此总操作次数为case1 + 1. **添加操作** `比如ab abc`
+2. 把s[1..i-1] 转化成t[1..j], 设操作数为case2， 我们只需要把s[i]从后面删除即可，总操作次数为case2 + 1. ** 删除操作 ** `比如abc ab`
+3. 如果我们可以把s[1..i-1]转化成t[1..j-1],设操作数为case3，如果s[i] == s[j],总操作数为case3， 否则需要替换一个字符，总操作数为case3 + 1. **替换操作** 比如`abc abe`
+
+我们取这三种情况的最小值作为最后的结果。
+
+我们使用`dp[i][j]`表示s[1..i]转化成t[1..j]的最小操作数。显然:
+
+* `dp[0][j] == j`, 只需要把t全部删除即可
+* 同理`dp[i][0] == i`
+
+因此该题转化成dp问题
+
+```c
+int minDistance(char *w1, char *w2)
+{
+	const int len1 = strlen(w1);
+	const int len2 = strlen(w2);
+	int dp[len1 + 1][len2 + 1];
+	dp[0][0] = 0;
+	for (int i = 1; i <= len1; ++i)
+		dp[i][0] = i;
+	for (int j = 1; j <= len2; ++j)
+		dp[0][j] = j;
+	for (int i = 1; i <= len1; ++i) {
+		for (int j = 1; j <= len2; ++j) {
+			int case1 = dp[i - 1][j] + 1;
+			int case2 = dp[i][j - 1] + 1;
+			int case3 = dp[i - 1][j - 1] + (w1[i - 1] == w2[j - 1] ? 0 : 1);
+			dp[i][j] = min3(case1, case2, case3);
+		}
+	}
+	return dp[len1][len2];
+}
+```

algorithms/EditDistance/in.txt

@@ -0,0 +1,5 @@
+aaa aaa
+bab aaa
+abcd defg
+cccccccccccccc c
+c cccccccccccccccc

algorithms/EditDistance/solve.c

@@ -0,0 +1,47 @@
+#include <stdio.h>
+#include <string.h>
+#include <stdlib.h>
+static inline int min2(int a, int b)
+{
+	return a < b ? a : b;
+}
+static inline int min3(int a, int b, int c)
+{
+	return min2(min2(a, b), c);
+}
+int minDistance(char *w1, char *w2)
+{
+	if (w1 == NULL || *w1 == 0) {
+		if (w2 == NULL || *w2 == 0)
+			return 0;
+		else
+			return strlen(w2);
+	}
+	if (w2 == NULL || *w2 == 0)
+		return strlen(w1);
+	const int len1 = strlen(w1);
+	const int len2 = strlen(w2);
+	int dp[len1 + 1][len2 + 1];
+	dp[0][0] = 0;
+	for (int i = 1; i <= len1; ++i)
+		dp[i][0] = i;
+	for (int j = 1; j <= len2; ++j)
+		dp[0][j] = j;
+	for (int i = 1; i <= len1; ++i) {
+		for (int j = 1; j <= len2; ++j) {
+			int case1 = dp[i - 1][j] + 1; // 删除当前字符
+			int case2 = dp[i][j - 1] + 1;
+			int case3 = dp[i - 1][j - 1] + (w1[i - 1] == w2[j - 1] ? 0 : 1);
+			dp[i][j] = min3(case1, case2, case3);
+		}
+	}
+	return dp[len1][len2];
+}
+int main(int argc, char **argv)
+{
+	char s1[20], s2[20];
+	while (scanf("%s%s", s1, s2) != EOF) {
+		printf("|%s, %s| = %d\n",s1, s2, minDistance(s1, s2));
+	}
+	return 0;
+}

database/DepartmentTopThreeSalaries/solve.sql

@@ -0,0 +1,18 @@
+use leetcode;
+drop table if exists Employee;
+create table Employee(Id int, Name varchar(20), Salary int, DepartmentId int);
+insert into Employee values
+(1, "Joe", 70000, 1),
+(2, "Henry", 80000, 2),
+(3,"Sam", 60000, 2),
+(4,"Max", 90000, 1),
+(5, "Janet", 69000, 1),
+(6, "Randy", 85000, 1);
+
+drop table if exists Department;
+create table Department(Id int, Name varchar(20));
+insert into Department values
+(1,"IT"),
+(2,"Sales");
+
+select top(Salary) from Employee group by DepartmentId;