力扣72题 编辑距离

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()];

    }

};