LeetCode 1143. Longest Common Subsequence #21
Description
Given two strings text1 and text2, return the length of their longest common subsequence.
A subsequence of a string is a new string generated from the original string with some characters(can be none) deleted without changing the relative order of the remaining characters. (eg, "ace" is a subsequence of "abcde" while "aec" is not). A common subsequence of two strings is a subsequence that is common to both strings.
If there is no common subsequence, return 0.
Example 1:
Input: text1 = "abcde", text2 = "ace"
Output: 3
Explanation: The longest common subsequence is "ace" and its length is 3.
Example 2:
Input: text1 = "abc", text2 = "abc"
Output: 3
Explanation: The longest common subsequence is "abc" and its length is 3.
Example 3:
Input: text1 = "abc", text2 = "def"
Output: 0
Explanation: There is no such common subsequence, so the result is 0.
Constraints:
- 1 <= text1.length <= 1000
- 1 <= text2.length <= 1000
- The input strings consist of lowercase English characters only.
仍是用线性DP的思路去做。创建一个二维dp数组，dp[i][j]代表第一个字符串的前 i个字符和第二个字符串的前 j个字符的最长公共子序列长度(从下标1开始)。
初始时：dp[i][0] = dp[0][j] = 0，表示空串和任何非空串都没有公共子序列。
状态转移方式：如果当前第一个字符串的第i个字符和第二个字符串的第j个字符相等，那么
1. dp[i][j] = dp[i - 1][j - 1] + 1
2. dp[i][j] = Math.max(dp[i - 1][j - 1] + 1, Math.max(dp[i - 1][j], dp[i][j - 1]))
否则
dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1])
返回dp[m][n]。
class Solution {
public int longestCommonSubsequence(String text1, String text2) {
int m = text1.length(), n = text2.length();
int[][] dp = new int[m + 1][n + 1];
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
if (text1.charAt(i - 1) == text2.charAt(j - 1)) {
// 1.dp[i][j] = dp[i - 1][j - 1] + 1;
// 2.
dp[i][j] = Math.max(dp[i - 1][j - 1] + 1,
Math.max(dp[i - 1][j], dp[i][j - 1]));
}
else {
dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
}
}
}
return dp[m][n];
}
}总结:
首先这是动态规划中的Min/Max paths问题，创建二维数组，设置大小((m + 1) * (n + 1))，这样就不用考虑边界问题，如果是(m* n)那么要初始化第一行第一列。确定dp[i][j]的含义——第一个字符串的前 i个字符和第二个字符串的前 j个字符的最长公共子序列长度。每个位置代表一种状态，接下来确定状态转移函数，dp[i][j]由其左边，上边和左上边的位置决定。当字符串1的第i个字符与字符串2的第j个字符不相等的话，可以忽视text1[i]或者text2[j]的状态，从左边和上边选择最大的子序列长度；反之如果字符相等，那么要从左上角的最大公共序列长度取出再加一。可以举一个例子，比如字符串1是abcd，另一个是aec，当i = j = 2时，其dp数值等于第1个索引也是第0个索引的值加一。
因为有m * n个状态，所以空间复杂度为O(m * n)，时间复杂度为O(m * n)。
参考资料: