Day44

1143.最长公共子序列

class Solution {
public:int longestCommonSubsequence(string text1, string text2) {vector<vector<int>> dp(text1.size() + 1, vector<int>(text2.size() + 1, 0));for (int i = 1; i <= text1.size(); i++) {for (int j = 1; j <= text2.size(); j++) {if (text1[i - 1] == text2[j - 1]) {dp[i][j] = dp[i - 1][j - 1] + 1;} else {dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);}}}return dp[text1.size()][text2.size()];}
};

五部曲

1.dp数组及下标定义：dp[i][j]：长度为[0, i - 1]的字符串text1与长度为[0, j - 1]的字符串text2的最长公共子序列为dp[i][j]

2.递推公式：dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);

3.初始化： 统一初始为0

4.遍历顺序：从前向后，从上到下

5.数组的数据应该是怎样的：