leetcode 72. Edit Distance
2016-09-30 21:42
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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
分析:
类似求最长公共子序列。
两个字符串,判断他们之间的编辑距离,可以通过三个操作,删除,添加,替换。每种操作都算距离加一。例如“ab”和“abc”的距离为1.
动态规划:用dis[i][j]记录string1的前i个和string2的前j个的距离。那么可以知道:
1.如果str1的第i个,也就是str1[i-1]和str2的第j个也就是str2[j-1]相等的话,那么
dis[i][j] = dis[i-1][j-1]
2.如果str[i-1] != str2[j-1]
2.1 通过替换操作把str[i-1]替换成str2[j-1],那么
dis[i][j] = dis[i-1][j-1] + 1;
2.2 通过插入操作在str1后面插入str2[j-1], 那么就相当于计算
dis[i][j] = dis[i][j-1] + 1;
2.3 通过插入操作在str2后面插入str1[i-1],那么就是
dis[i][j] = dis[i-1][j] + 1;
在上述三个中选一个最小的。迭代更新。
ac代码:
class Solution {
public:
int minDistance(string word1, string word2) {
int i,j,L1=word1.length(),L2=word2.length();
if(L1==0) return L2;
if(L2==0) return L1;
int dis[L1+1][L2+1];
for(i=0;i<=L1;i++)
dis[i][0]=i;
for(j=0;j<=L2;j++)
dis[0][j]=j;
for(i=1;i<=L1;i++)
{
for(j=1;j<=L2;j++)
{
if(word1[i-1]!=word2[j-1])
{
dis[i][j]=min(dis[i-1][j]+1,dis[i][j-1]+1);
dis[i][j]=min(dis[i-1][j-1]+1,dis[i][j]);
}
else
{
dis[i][j]=dis[i-1][j-1];
}
}
}
return dis[L1][L2];
}
};
You have the following 3 operations permitted on a word:
a) Insert a character
b) Delete a character
c) Replace a character
分析:
类似求最长公共子序列。
两个字符串,判断他们之间的编辑距离,可以通过三个操作,删除,添加,替换。每种操作都算距离加一。例如“ab”和“abc”的距离为1.
动态规划:用dis[i][j]记录string1的前i个和string2的前j个的距离。那么可以知道:
1.如果str1的第i个,也就是str1[i-1]和str2的第j个也就是str2[j-1]相等的话,那么
dis[i][j] = dis[i-1][j-1]
2.如果str[i-1] != str2[j-1]
2.1 通过替换操作把str[i-1]替换成str2[j-1],那么
dis[i][j] = dis[i-1][j-1] + 1;
2.2 通过插入操作在str1后面插入str2[j-1], 那么就相当于计算
dis[i][j] = dis[i][j-1] + 1;
2.3 通过插入操作在str2后面插入str1[i-1],那么就是
dis[i][j] = dis[i-1][j] + 1;
在上述三个中选一个最小的。迭代更新。
ac代码:
class Solution {
public:
int minDistance(string word1, string word2) {
int i,j,L1=word1.length(),L2=word2.length();
if(L1==0) return L2;
if(L2==0) return L1;
int dis[L1+1][L2+1];
for(i=0;i<=L1;i++)
dis[i][0]=i;
for(j=0;j<=L2;j++)
dis[0][j]=j;
for(i=1;i<=L1;i++)
{
for(j=1;j<=L2;j++)
{
if(word1[i-1]!=word2[j-1])
{
dis[i][j]=min(dis[i-1][j]+1,dis[i][j-1]+1);
dis[i][j]=min(dis[i-1][j-1]+1,dis[i][j]);
}
else
{
dis[i][j]=dis[i-1][j-1];
}
}
}
return dis[L1][L2];
}
};
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