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LeetCode583.两个字符串的删除操作
题目链接:https://leetcode.cn/problems/delete-operation-for-two-strings/
思路:dp[i][j]:使以i-1为尾的word1和以j-1为尾的word2相同的最小步数
当word1[i - 1] == word2[j - 1]时,word1[i - 1]和word2[j - 1]都不用删除,dp[i][j]就等于dp[i - 1][j - 1].
其他情况时,dp[i][j]就应该取dp[i - 1][j] + 1和dp[i][j - 1] + 1中的最小值,这里为什么加1?因为word1[i - 1]不等于word2[j - 1]的时候,就要删除再分别从两个角度考虑。
动规:
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 = 1; 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] + 1, dp[i][j - 1] + 1); }}return dp[word1.size()][word2.size()];}
};
LeetCode72.编辑距离
题目链接:https://leetcode.cn/problems/edit-distance/
思路:不多说了,你就画吧。dp数组一画一个不吱声。
动规:
class Solution {
public:int minDistance(string word1, string word2) {vector<vector<int>> dp(word2.size() + 1, vector<int>(word1.size() + 1, 0));for(int i = 1; i <= word2.size(); i++) dp[i][0] = i;for(int j = 1; j <= word1.size(); j++) dp[0][j] = j;for(int i = 1; i <= word2.size(); i++) {for(int j = 1; j <= word1.size(); j++) {if(word1[j - 1] == word2[i - 1]) dp[i][j] = dp[i - 1][j - 1];else dp[i][j] = min(dp[i - 1][j - 1] + 1, min(dp[i - 1][j] + 1, dp[i][j - 1] + 1));}}return dp[word2.size()][word1.size()];}
};
总结:被我找到窍门了,序列问题dp的关键在于将二维数组画出来,画的过程中自己一步一步推导每一个值的由来,这样在写状态转移方程的时候就会简单很多。