给定小写字母的字符串str ，将给定的字符串分成尽可能多的子字符串，以使给定字符串中的每个字符出现在单个子字符串中。任务是打印所有此类分区的长度。
例子：
Input: str = “acbbcc”
Output: 1 5
Explanation:
Possible partitions where each character of the strings occurs in at most one partition are “a” and “cbbcc”.
Therefore, the length is {1, 5}
Input: str = “abaccbdeffed”
Output: 6 6
Explanation:
Possible partitions where each character of the strings occurs in at most one partition are “abaccb” and “deffed”.
Therefore, the length is {6, 6}
方法：使用贪婪方法可以轻松解决此问题。请按照以下步骤解决问题。
- 将所有字符的最后一个索引存储在字符串。
- 由于字符串仅包含小写字母，因此只需使用固定大小为26的数组来存储每个字符的最后一个索引。
- 遍历给定的字符串，然后执行以下步骤：
- 当前字符添加到该分区如果字符的最后一个位置超过了当前的指数和增加分区的长度。
- 如果当前字符的最后一个索引等于当前索引，则存储其长度并继续到下一个字符，依此类推。
- 完成上述步骤后，打印所有存储的长度。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the length of
// all partitions oof a string such
// that each characters occurs in
// a single substring
void partitionString(string s)
{
int n = s.size();
// Stores last index of string s
vector ans;
if (n == 0) {
cout << "-1";
return;
}
// Find the last position of
// each letter in the string
vector last_pos(26, -1);
for (int i = n - 1; i >= 0; --i) {
// Update the last index
if (last_pos[s[i] - 'a'] == -1) {
last_pos[s[i] - 'a'] = i;
}
}
int minp = -1, plen = 0;
// Iterate the given string
for (int i = 0; i < n; ++i) {
// Get the last index of s[i]
int lp = last_pos[s[i] - 'a'];
// Extend the current partition
// characters last pos
minp = max(minp, lp);
// Increase len of partition
++plen;
// if the current pos of
// character equals the min pos
// then the end of partition
if (i == minp) {
// Store the length
ans.push_back(plen);
minp = -1;
plen = 0;
}
}
// Print all the partition lengths
for (int i = 0;
i < (int)ans.size(); i++) {
cout << ans[i] << " ";
}
}
// Driver Code
int main()
{
// Given string str
string str = "acbbcc";
// Function Call
partitionString(str);
return 0;
} Java
// Java program for
// the above approach
import java.util.*;
class GFG{
// Function to find the length of
// all partitions oof a String such
// that each characters occurs in
// a single subString
static void partitionString(String s)
{
int n = s.length();
// Stores last index of String s
Vector ans =
new Vector();
if (n == 0)
{
System.out.print("-1");
return;
}
// Find the last position of
// each letter in the String
int []last_pos = new int[26];
Arrays.fill(last_pos, -1);
for (int i = n - 1; i >= 0; --i)
{
// Update the last index
if (last_pos[s.charAt(i) - 'a'] == -1)
{
last_pos[s.charAt(i) - 'a'] = i;
}
}
int minp = -1, plen = 0;
// Iterate the given String
for (int i = 0; i < n; ++i)
{
// Get the last index of s[i]
int lp = last_pos[s.charAt(i) - 'a'];
// Extend the current partition
// characters last pos
minp = Math.max(minp, lp);
// Increase len of partition
++plen;
// if the current pos of
// character equals the min pos
// then the end of partition
if (i == minp)
{
// Store the length
ans.add(plen);
minp = -1;
plen = 0;
}
}
// Print all the partition lengths
for (int i = 0; i < (int)ans.size(); i++)
{
System.out.print(ans.get(i) + " ");
}
}
// Driver Code
public static void main(String[] args)
{
// Given String str
String str = "acbbcc";
// Function Call
partitionString(str);
}
}
// This code is contributed by 29AjayKumar Python3
# Python3 program for the above approach
# Function to find the length of
# all partitions oof a string such
# that each characters occurs in
# a single substring
def partitionString(s):
n = len(s)
# Stores last index of string s
ans = []
if (n == 0):
print("-1")
return
# Find the last position of
# each letter in the string
last_pos = [-1] * 26
for i in range(n - 1 , -1 , -1):
# Update the last index
if (last_pos[ord(s[i]) - ord('a')] == -1):
last_pos[ord(s[i]) - ord('a')] = i
minp = -1
plen = 0
# Iterate the given string
for i in range(n):
# Get the last index of s[i]
lp = last_pos[ord(s[i]) - ord('a')]
# Extend the current partition
# characters last pos
minp = max(minp, lp)
# Increase len of partition
plen += 1
# if the current pos of
# character equals the min pos
# then the end of partition
if (i == minp):
# Store the length
ans.append(plen)
minp = -1
plen = 0
# Print all the partition lengths
for i in range(len(ans)):
print(ans[i], end = " ")
# Driver Code
# Given string str
str = "acbbcc"
# Function call
partitionString(str)
# This code is contributed by code_huntC#
// C# program for
// the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the length of
// all partitions oof a String such
// that each characters occurs in
// a single subString
static void partitionString(String s)
{
int n = s.Length;
// Stores last index of String s
List ans = new List();
if (n == 0)
{
Console.Write("-1");
return;
}
// Find the last position of
// each letter in the String
int []last_pos = new int[26];
for (int i = 0; i < 26; ++i)
{
last_pos[i] = -1;
}
for (int i = n - 1; i >= 0; --i)
{
// Update the last index
if (last_pos[s[i] - 'a'] == -1)
{
last_pos[s[i] - 'a'] = i;
}
}
int minp = -1, plen = 0;
// Iterate the given String
for (int i = 0; i < n; ++i)
{
// Get the last index of s[i]
int lp = last_pos[s[i] - 'a'];
// Extend the current partition
// characters last pos
minp = Math.Max(minp, lp);
// Increase len of partition
++plen;
// if the current pos of
// character equals the min pos
// then the end of partition
if (i == minp)
{
// Store the length
ans.Add(plen);
minp = -1;
plen = 0;
}
}
// Print all the partition lengths
for (int i = 0; i < (int)ans.Count; i++)
{
Console.Write(ans[i] + " ");
}
}
// Driver Code
public static void Main(String[] args)
{
// Given String str
String str = "acbbcc";
// Function Call
partitionString(str);
}
}
// This code is contributed by Rajput-Ji 输出：
1 5
时间复杂度： O(N)
辅助空间： O(N)