题型：动态规划，难度大
1.确定dp数组以及下标的含义
dp[i][j] 表示以下标i-1为结尾的字符串word1，和以下标j-1为结尾的字符串word2，最近编辑距离为dp[i][j]。
2.确定递推公式
class Solution { public:int minDistance(string word1, string word2) {vector<vector<int>> dp(word1.size() + 1, vector<int>(word2.size() + 1, 0));for (int i = 0; i <= word1.size(); i++) dp[i][0] = i;for (int j = 0; 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 - 1], dp[i - 1][j], dp[i][j - 1]}) + 1;}}}return dp[word1.size()][word2.size()];} };
3.dp数组如何初始化
dp[i][j]表示以下标i-1为结尾的字符串word1，和下标j-1为结尾的字符串word2，最近编辑距离为dp[i][j]
dp[i][0]：以下标i-1为结尾的字符串word1，和空字符串word2，最近编辑距离为dp[i][0]。
那么dp[i][0]就应该是i，对word1里的元素全部做删除操作，即：dp[i][0]=i；
同理dp[0][j]=j;
for (int i = 0; i <= word1.size(); i++) dp[i][0] = i; for (int j = 0; j <= word2.size(); j++) dp[0][j] = j;
4.遍历顺序
dp[i][j]=dp[i-1][j]
dp[i][j]=dp[i-1][j-1]+1
dp[i][j]=dp[i][j-1]+1
dp[i][j]dp[i-1][j]+1
dp矩阵中一定是从左到右从上到下去遍历
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 - 1], dp[i - 1][j], dp[i][j - 1]}) + 1;}}
}
5.举例推导dp数组
class Solution {
public:
int minDistance(string word1, string word2) {
vector<vector<int>>dp(word1.size()+1,vector<int>(word2.size()+1,0));
for(int i=0;i<=word1.size();i++) dp[i][0] = i;
for(int j=0;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-1],dp[i-1][j],dp[i][j-1]})+1;
}
}
}
return dp[word1.size()][word2.size()];
}
};