题目链接:583. 两个字符串的删除操作
1、dp数组及下标含义:
dp[i][j]表示下标为i-1之前的word1字符串和下标为j-1之前的word2字符串相同所需删除字符的最小操作步骤。
2、递推公式:
if(word2[j - 1] == word1[i - 1]) dp[i][j] = dp[i - 1][j - 1];
else {
dp[i][j] = min(dp[i - 1][j] + 1, dp[i][j - 1] + 1);
}
3、初始化:
for(int i = 0; i <= word1.size(); i++) dp[i][0] = i;
for(int j = 1; j <= word2.size(); j++) dp[0][j] = j;
4、遍历顺序:
由递推公式可知,dp[i][j]依赖于dp[i-1][j]、dp[i][j-1]和dp[i-1][j-1]。
5、打印dp数组
代码如下:
class Solution {
public:
int minDistance(string word1, string word2) {
vector<vector<int>>dp(word1.size() + 1, vector<int>(word2.size() + 1));
for(int i = 0; i <= word1.size(); i++) dp[i][0] = i;
for(int j = 1; j <= word2.size(); j++) dp[0][j] = j;
for(int i = 1; i <= word1.size(); i++) {
for(int j = 1; j <= word2.size(); j++) {
if(word2[j - 1] == word1[i - 1]) dp[i][j] = dp[i - 1][j - 1];
else {
dp[i][j] = min(dp[i - 1][j] + 1, dp[i][j - 1] + 1);
}
}
}
return dp[word1.size()][word2.size()];
}
};
题目链接:72. 编辑距离
1、dp数组及其下标含义:
dp[i][j]表示下标i-1之前的字符串word1通过增、删、替换,改变成下标为j-1之前的字符串word2所需的最少步骤。
2、递推公式:
if(word1[i - 1] == word2[j - 1]) dp[i][j] = dp[i - 1][j - 1];
else dp[i][j] = min(dp[i - 1][j], min(dp[i][j - 1], dp[i - 1][j - 1])) + 1;
3、初始化:
for(int i = 0; i <= word1.size(); i++) dp[i][0] = i;
for(int j = 1; j <= word2.size(); j++) dp[0][j] = j;
4、遍历顺序:
从上到下,从左到右。
5、打印dp数组;
代码如下:
class Solution {
public:
int minDistance(string word1, string word2) {
vector<vector<int>>dp(word1.size() + 1, vector<int>(word2.size() + 1));
for(int i = 0; i <= word1.size(); i++) dp[i][0] = i;
for(int j = 1; j <= word2.size(); j++) dp[0][j] = j;
for(int i = 1; i <= word1.size(); i++) {
for(int j = 1; j <= word2.size(); j++) {
if(word1[i - 1] == word2[j - 1]) dp[i][j] = dp[i - 1][j - 1];
else dp[i][j] = min(dp[i - 1][j], min(dp[i][j - 1], dp[i - 1][j - 1])) + 1;
}
}
return dp[word1.size()][word2.size()];
}
};