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[LeetCode]Edit Distance

2015-11-26 15:36 676 查看
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

动态规划。f(i,j)代表 word1 前i位 和 word2 前 j位 转换时最小的次数。

分两种情况,如果word1[i] == word2 [j]:

i位和j位相等,说明 f(i,j) = f(i-1,j-1).没有cost

如果word1[i] 和 word2[j] 不相等,分三种情况:

删除 f(i-1,j) 替换 f(i-1,j-1)
插入 f(i,j-1) f(i,j) = min(f(i-1,j),f(i-1,j-1),f(i,j-1))+1;

class Solution {
public:
int minDistance(string word1, string word2) {
int len1 = word1.size();
int len2 = word2.size();
if(len1==0)
return len2;
if(len2==0)
return len1;
vector<vector<int>> Dp (len1+1,vector<int>(len2+1,0)) ;
for(int i=0; i<len1+1; ++i){
Dp[i][0] = i;
}
for(int i=0; i<len2+1; ++i){
Dp[0][i] = i;
}
for(int i=1; i<len1+1; ++i){
for(int j=1; j<len2+1; ++j){
if(word1[i-1] == word2[j-1]){
Dp[i][j] = Dp[i-1][j-1];
}
else{
Dp[i][j] = 1+min(Dp[i-1][j-1],Dp[i-1][j],Dp[i][j-1]);
}
}
}
return Dp[len1][len2];
}
int min(int a,int b, int c){
int min = a;
if(min>b)
min = b;
if(min>c)
min = c;
return min;
}
};
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