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The longest common substring title 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.
Example 1
Input "1AB2345CD", "12345EF"
returns the value "2345"
The longest common substring code:
import java.util.*;
public class Solution {
/**
* longest common substring
* @param str1 string字符串 the string
* @param str2 string字符串 the string
* @return string字符串
*/
public String LCS (String str1, String str2) {
if(str1 == null || str2 == null || str1.equals("") || str2.equals("")){
return "-1";
}
int indexMax = 0;
int maxLen = 0;
int m = str1.length();
int n = str2.length();
//dp[i][j]代表 str1[0~i-1]和str2[0~j-1] 的最长公共子串的长度
int[][] dp = new int[m][n];
for(int i = 0; i < m; ++ i){
for(int j = 0; j < n; ++j){
if(str1.charAt(i) == str2.charAt(j)){
if(i == 0 || j == 0){
dp[i][j] = 1;
}else{
dp[i][j] = dp[i - 1][j - 1] + 1;
}
}//else 是str1[i]!=str2[j]的情况,这种情况下dp[i][j]=0,由于初始化已经将其设置为0,所以这里不再写。
//处理完dp[i][j]之后,查看一下是否需要记录下来
if(maxLen < dp[i][j]){
maxLen = dp[i][j]; //记录下最长公共子串的长度
indexMax = i; //记录下出现“最长公共子串”时的末尾字符的位置
}
}
}
if(maxLen == 0) return "-1";
//字符串截取的长度有(end-start+1) = maxLen, 那么start = indexMax +1-maxLen
// maxLen即为所截取的字符串的长度。
return str1.substring(indexMax - maxLen + 1 , indexMax + 1);
}
}
The longest common subsequence title description
Given two strings str1 and str2, output the longest common subsequence of a string. If the longest common subsequence is empty, -1 is output.
Example 1
Input "1A2C3D4B56", "B1D23CA45B6A" and the
return value "123456"
shows that "123456" and "12C4B6" are the longest common subsequences, and any one is output.
Longest common subsequence code
import java.util.*;
public class Solution {
/**
* longest common subsequence
* @param s1 string字符串 the string
* @param s2 string字符串 the string
* @return string字符串
*/
public String LCS (String s1, String s2) {
// write code here
int str1Len = s1.length();
int str2Len = s2.length();
int[][] cLenNUm = new int[s1.length() + 1][s2.length() + 1];//默认赋值，[0][?],[?][0]默认两侧皆0,类似公式中0的场景
//构造一个LCS长度数组
for (int i = 1; i <= str1Len; i++) {
for (int j = 1; j <= str2Len; j++) {
if (s1.charAt(i - 1) == s2.charAt(j - 1)) {//对应公式第二条相等
cLenNUm[i][j] = cLenNUm[i - 1][j - 1] + 1;
} else {//对应公式第三条不相等
cLenNUm[i][j] = Math.max(cLenNUm[i][j - 1], cLenNUm[i - 1][j]);
}
}
}
//反推结果
int i = str1Len;
int j = str2Len;
StringBuffer sb = new StringBuffer();//作为结果
while (i > 0 && j > 0) {//这里其实处理了i=0,j=0的，对应公式0的反推场景
if (s1.charAt(i - 1) == s2.charAt(j - 1)) {//反推公式中不相等的场景
//该值一定是被选取到的，根据之前的公式，知道两条字符串的下标都前进一位
sb.append(s1.charAt(i - 1));
i--;
j--;
} else {//对应公式中不相等的反推场景
if (cLenNUm[i][j - 1] > cLenNUm[i - 1][j]) {//找大的那个方向，此处是左边大于上面，则该处的结果是来自左边
j--;
} else if (cLenNUm[i][j - 1] < cLenNUm[i - 1][j]) {
i--;
} else if (cLenNUm[i][j - 1] == cLenNUm[i - 1][j]) {
//对于有分支的可能时，我们选取单方向
j--; //此结果对于结果1所选取方向，str1的下标左移一位.替换为j--，则结果对应与结果2选取的方向
}
}
}
//由于是从后往前加入字符的，需要反转才能得到正确结果
return sb.reverse().toString();
}
}
Both questions are dynamic programming
The longest common substring: the letters in the substring required here areThere is no space between other characters
int[][] dp = new int[len1+1][len2+1];
dp[0][0] = 0;
int maxlen = Integer.MIN_VALUE, indexmax = -1;
for(int i=0;i<len1;++i){
for(int j=0;j<len2;++j){
if(str1.charAt(i) == str2.charAt(j)){
if(i == 0 ||j == 0) dp[i][j] = 1;
else dp[i][j] = dp[i-1][j-1] + 1;
}
if(maxlen < dp[i][j]){
maxlen = dp[i][j];
indexmax = i;
}
}
}
The longest subsequence, here in the common substringCan space other characters
int[][] dp = new int[len1+1][len2+1];
dp[0][0] = 0;
for(int i=1;i<=len1;++i){
for(int j=1;j<=len2;++j){
if(s1.charAt(i-1) == s2.charAt(j-1)){
dp[i][j] = dp[i-1][j-1]+1;
}else{
dp[i][j] = Math.max(dp[i-1][j], dp[i][j-1]);
}
}
}