给定两个字符串str1和str2 ，其中str2是str1的子序列，任务是找到str1的最长子字符串的长度，当删除时，使字符串str2和str1相等。
例子：
Input: str1 = “programmingbloods”, str2 = “ibloods”
Output: 8
Explanation:
Substrings to be removed from str1 are [“programm”, “ng”]. Therefore, the length of the longest deleted substring is 8.
Input: str1=“GeeksforGeeks”, str2=“forks”
Output: 5
朴素的方法：最简单的方法是生成str1 的所有可能子字符串，对于每个子字符串，将其从str1 中删除并检查结果字符串是否等于str2 。打印最长的此类子串的长度。
时间复杂度： O(2 N )
辅助空间： O(N)
高效的方法：为了优化上述方法，思想是在str1 中找到str2的出现。从左到右遍历两个字符串并检查两个字符是否相等。如果发现是真的，则在两个字符串向右进行。否则，仅在str2 中向右继续。同样，找到STR1 STR2最后一次出现，遍历从右到左字符串和同样进行。请按照以下步骤解决问题：
- 初始化一个变量， res=0存放最长删除子串的长度。
- 创建一个数组pos[]来存储str2在str1 中第一次出现的位置。
- 从左到右遍历两个字符串，通过删除str1的一些字符来存储str2第一次出现的位置。
- 初始化一个变量lastPos = length(str1)-1通过删除str1 的一些字符来存储当前字符在str2的最后一次出现中的位置。
- 从右到左遍历两个字符串并检查两个字符匹配，然后在pos[] 中找到str2的当前字符的位置，否则继续。
- 如果res > ( lastPos – pos[i-1]-1)则更新res = lastPos – pos[i-1]-1。
- 最后，返回res 。
下面是上述方法的实现：
C++
// C++ Program to implement
// the above approach
#include
using namespace std;
// Function to print the length
// of longest substring to be deleted
int longDelSub(string str1, string str2)
{
// Stores the length of string
int N = str1.size();
int M = str2.size();
// Store the position of
// previous matched
// character of str1
int prev_pos = 0;
// Store the position of first
// occurrence of str2 in str1
int pos[M];
// Find the position of the
// first occurrence of str2
for (int i = 0; i < M; i++) {
// Store the index of str1
int index = prev_pos;
// If both characters not matched
while (index < N
&& str1[index] != str2[i]) {
index++;
}
pos[i] = index;
prev_pos = index + 1;
}
// Store the length of the
// longest deleted substring
int res = N - prev_pos;
prev_pos = N - 1;
// Store the position of last
// occurrence of str2 in str1
for (int i = M - 1; i >= 0; i--) {
int index = prev_pos;
// If both characters not matched
while (index >= 0
&& str1[index] != str2[i]) {
index--;
}
// Update res
if (i != 0) {
res = max(
res,
index - pos[i - 1] - 1);
}
prev_pos = index - 1;
}
// Update res.
res = max(res, prev_pos + 1);
return res;
}
// Driver Code
int main()
{
// Given string
string str1 = "GeeksforGeeks";
string str2 = "forks";
// Function Call
cout << longDelSub(str1, str2);
return 0;
} Java
// Java program to implement
// the above approach
import java.io.*;
class GFG{
// Function to print the length
// of longest substring to be deleted
static int longDelSub(String str1, String str2)
{
// Stores the length of string
int N = str1.length();
int M = str2.length();
// Store the position of
// previous matched
// character of str1
int prev_pos = 0;
// Store the position of first
// occurrence of str2 in str1
int pos[] = new int[M];
// Find the position of the
// first occurrence of str2
for(int i = 0; i < M; i++)
{
// Store the index of str1
int index = prev_pos;
// If both characters not matched
while (index < N &&
str1.charAt(index) !=
str2.charAt(i))
{
index++;
}
pos[i] = index;
prev_pos = index + 1;
}
// Store the length of the
// longest deleted substring
int res = N - prev_pos;
prev_pos = N - 1;
// Store the position of last
// occurrence of str2 in str1
for(int i = M - 1; i >= 0; i--)
{
int index = prev_pos;
// If both characters not matched
while (index >= 0 &&
str1.charAt(index) !=
str2.charAt(i))
{
index--;
}
// Update res
if (i != 0)
{
res = Math.max(res,
index -
pos[i - 1] - 1);
}
prev_pos = index - 1;
}
// Update res.
res = Math.max(res, prev_pos + 1);
return res;
}
// Driver Code
public static void main (String[] args)
{
// Given string
String str1 = "GeeksforGeeks";
String str2 = "forks";
// Function call
System.out.print(longDelSub(str1, str2));
}
}
// This code is contributed by code_huntPython3
# Python3 program to implement
# the above approach
# Function to prthe length
# of longest substring to be deleted
def longDelSub(str1, str2):
# Stores the length of string
N = len(str1)
M = len(str2)
# Store the position of
# previous matched
# character of str1
prev_pos = 0
# Store the position of first
# occurrence of str2 in str1
pos = [0] * M
# Find the position of the
# first occurrence of str2
for i in range(M):
# Store the index of str1
index = prev_pos
# If both characters not matched
while (index < N and
str1[index] != str2[i]):
index += 1
pos[i] = index
prev_pos = index + 1
# Store the length of the
# longest deleted substring
res = N - prev_pos
prev_pos = N - 1
# Store the position of last
# occurrence of str2 in str1
for i in range(M - 1, -1, -1):
index = prev_pos
# If both characters not matched
while (index >= 0 and
str1[index] != str2[i]):
index -= 1
# Update res
if (i != 0) :
res = max(res,
index -
pos[i - 1] - 1)
prev_pos = index - 1
# Update res.
res = max(res, prev_pos + 1)
return res
# Driver Code
# Given string
str1 = "GeeksforGeeks"
str2 = "forks"
# Function call
print(longDelSub(str1, str2))
# This code is contributed by code_huntC#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to print the length
// of longest substring to be deleted
static int longDelSub(string str1,
string str2)
{
// Stores the length of string
int N = str1.Length;
int M = str2.Length;
// Store the position of
// previous matched
// character of str1
int prev_pos = 0;
// Store the position of first
// occurrence of str2 in str1
int[] pos = new int[M];
// Find the position of the
// first occurrence of str2
for(int i = 0; i < M; i++)
{
// Store the index of str1
int index = prev_pos;
// If both characters not matched
while (index < N &&
str1[index] != str2[i])
{
index++;
}
pos[i] = index;
prev_pos = index + 1;
}
// Store the length of the
// longest deleted substring
int res = N - prev_pos;
prev_pos = N - 1;
// Store the position of last
// occurrence of str2 in str1
for(int i = M - 1; i >= 0; i--)
{
int index = prev_pos;
// If both characters not matched
while (index >= 0 &&
str1[index] != str2[i])
{
index--;
}
// Update res
if (i != 0)
{
res = Math.Max(res,
index -
pos[i - 1] - 1);
}
prev_pos = index - 1;
}
// Update res.
res = Math.Max(res, prev_pos + 1);
return res;
}
// Driver code
public static void Main()
{
// Given string
string str1 = "GeeksforGeeks";
string str2 = "forks";
// Function call
Console.Write(longDelSub(str1, str2));
}
}
// This code is contributed by code_hunt输出：
5
时间复杂度： O(N + M)
辅助空间： O(M)
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