Problem Description
Given two strings str1 and str2, output the longest common substring of the two strings. If the longest common substring is empty, output -1.
Input description:
input two strings
Output description:
output the longest common substring
Example
Example 1
Enter
"1AB2345CD", "12345EF"
Output
"2345"
Solutions
analysis
- The longest common subsequence (Longest Common Subsequence, LCS for short) is a very classic interview question, because its solution is a typical two-dimensional dynamic programming, and most of the more difficult string problems follow this question.
- Method introduction
(1) The definition of dp array: int[][] dp = new int[m + 1][n + 1]; The value of the array is the length of the common string
(2) The assignment of a two-dimensional array:
(3 ) It can be seen from the chart that the length formula of the longest common string is: dp[i + 1][j + 1] = dp[i][j] + 1
method
- Use two-dimensional dynamic programming + some optimizations.
Code
// 思路1
public class Solution {
public String LCS (String str1, String str2) {
// write code here
int m = str1.length(), n = str2.length();
int[][] dp = new int[m + 1][n + 1];
// 记录最长公共字串的长度
int maxLen = 0;
int index = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
// 获取两个串字符
char c1 = str1.charAt(i), c2 = str2.charAt(j);
if (c1 == c2) {
// 运用二维动态规划总结的公式
dp[i + 1][j + 1] = dp[i][j] + 1;
if (dp[i + 1][j + 1] > maxLen) {
maxLen = dp[i + 1][j + 1];
// 记录最长公共字串结束的索引
index = j + 1;
}
}
}
}
if (maxLen == 0) {
return "";
}
return str2.substring(index - maxLen, index);
}
}
Time complexity analysis:
O(MN): Two-layer loop, traversing the string, so the time complexity is the product of the length of the string.
Space complexity analysis:
O(MN): An additional two-dimensional array is used to store the length of the longest common string, (m+1)(n+1) = mn + m + n + 1, m and n can be ignored The law is constant, so the space complexity is MN.
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