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:只能将字符串分成更小的区间字符串来考虑,从下向上分析。那么需要使用动态规划的方法。
设状态为f[i][j],表示A[0,i]、B[0,j]之间的最小编辑距离。设A[0,i]和B[0,j]的形式为str1c和str2d。
1、如果c==d,那么f[i][j]=f[i-1][j-1]。
2、如果c!=d:
(1)如果将c换成d,那么f[i][j]=f[i-1][j-1]+1;
(2)如果将c删除,那么f[i][j]=f[i-1][j]+1;
(3)如果将d删除,那么f[i][j]=f[i[j-1]+1;
/*二维动态规划代码*/
class Solution { public: int minDistance(string word1, string word2) { int f[word1.size()+1][word2.size()+1]; int i,j; for(i=0;i<=word1.size();i++) f[i][0]=i; for(j=0;j<=word2.size();j++) f[0][j]=j; for(int i=1;i<word1.size();i++) for(int j=1;i<word2.size();j++) { if(word1[i-1]==word2[j-1]) f[i][j]=f[i-1][j-1]; else{ int mn=min(f[i][j-1],f[i-1][j]); f[i][j]=i+min(mn,f[i-1][j-1]); } } return f[word1.size()][word2.size()]; } };
但是,在输入规模很大时,超时。
那么用滚动数组优化一下,节省空间。
</pre><p><pre name="code" class="cpp">class Solution { public: int minDistance(string word1, string word2) { int f[word2.size()+1]; int upper_left,temp; for(int i=0;i<=word2.size();++i) f[i]=i; for(int i=1;i<=word1.size();i++){ upper_left=f[0]; f[0]=i; for(int j=1;j<=word2.size();j++) { if(word1[i-1]==word2[j-1]){ temp=f[j]; f[j]=upper_left; upper_left=temp; } else{ // upper_left相当于 f[i-1][j-1],f[j-1]相当于f[i][j-1] int mn=min(upper_left,f[j-1]); temp=f[j]; f[j]=1+min(f[j],mn); //f[j]相当于f[i-1][j] upper_left=temp; } } } return f[word2.size()]; } };
学习:https://gitcafe.com/soulmachine/LeetCode