【LeetCode】72. 编辑距离
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2022-07-12 08:54:11
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给定两个单词 word1 和 word2,计算出将 word1 转换成 word2 所使用的最少操作数 。
你可以对一个单词进行如下三种操作:
插入一个字符
删除一个字符
替换一个字符
示例 1:
输入: word1 = "horse", word2 = "ros"
输出: 3
解释:
horse -> rorse (将 'h' 替换为 'r')
rorse -> rose (删除 'r')
rose -> ros (删除 'e')
示例 2:
输入: word1 = "intention", word2 = "execution"
输出: 5
解释:
intention -> inention (删除 't')
inention -> enention (将 'i' 替换为 'e')
enention -> exention (将 'n' 替换为 'x')
exention -> exection (将 'n' 替换为 'c')
exection -> execution (插入 'u')
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/edit-distance
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
动态规划二维数组存储
class Solution {
public:
int minDistance(string word1, string word2) {
vector<vector<int>> ans(word1.size()+1,vector<int>(word2.size()+1));
for(int i=0;i<=word2.size();i++)ans[0][i]=i;
for(int i=0;i<=word1.size();i++)ans[i][0]=i;
for(int i=1;i<=word1.size();i++){
for(int j=1;j<=word2.size();j++){
ans[i][j]=min(ans[i][j-1],ans[i-1][j])+1;
ans[i][j]=min(ans[i][j],ans[i-1][j-1]+(word1[i-1]==word2[j-1]?0:1));
}
}return ans[word1.size()][word2.size()];
}
};
动态规划,优化为一维数组存储
class Solution {
public:
int minDistance(string word1, string word2) {
// vector<vector<int>> ans(word1.size()+1,vector<int>(word2.size()+1));
// for(int i=0;i<=word2.size();i++)ans[0][i]=i;
// for(int i=0;i<=word1.size();i++)ans[i][0]=i;
vector<int>vv(word2.size()+1);
for(int i=0;i<=word2.size();i++)vv[i]=i;
int pre=0;
for(int i=1;i<=word1.size();i++){
vv[0]=i;
pre=i-1;
for(int j=1;j<=word2.size();j++){
int t=vv[j];
// ans[i][j]=min(ans[i][j-1],ans[i-1][j])+1;
// ans[i][j]=min(ans[i][j],ans[i-1][j-1]+(word1[i-1]==word2[j-1]?0:1));
vv[j]=min(vv[j-1],vv[j])+1;
vv[j]=min(vv[j],pre+(word1[i-1]==word2[j-1]?0:1));
if(word1[i-1]==word2[j-1])cout<<"["<<i-1<<","<<j-1<<pre<<"]";
cout<<vv[j];
pre=t;
}cout<<endl;
}return vv[word2.size()];
}
};
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