给定长度为N的字符串str 。任务是找出最多仅允许一次交换的字典上最小的字符串。也就是说，可以选择并交换两个索引1 <= i，j <= n 。此操作最多可以执行一次。
例子：
Input: str = “string”
Output: gtrins
Explanation:
Choose i=1, j=6, string becomes – gtrins. This is lexicographically smallest strings that can be formed.
Input: str = “zyxw”
Output: wyxz
方法：我们的想法是使用排序并计算给定字符串的最小辞书字符串可能。计算排序后的字符串，从给定的字符串找到第一个不匹配的字符，并将其替换为排序后的字符串最后一个不匹配的字符。
例如，让str =“ geeks”，而sorted =“ eegks”。第一个无与伦比的字符在首位。这和字符来交换，使得这个字符与排序字符串的字符相匹配。结果字典最小字符串。将“ g”替换为最后出现的“ e”时，字符串将变成字典上最小的eegks。
下面是上述方法的实现：
C++
// C++ implementation of the above approach
#include
using namespace std;
// Function to return the lexicographically
// smallest string that can be formed by
// swapping at most one character.
// The characters might not necessarily
// be adjacent.
string findSmallest(string s)
{
int len = s.size();
// Store last occurrence of every character
int loccur[26];
// Set -1 as default for every character.
memset(loccur, -1, sizeof(loccur));
for (int i = len - 1; i >= 0; --i) {
// Character index to fill
// in the last occurrence array
int chI = s[i] - 'a';
if (loccur[chI] == -1) {
// If this is true then this
// character is being visited
// for the first time from the last
// Thus last occurrence of this
// character is stored in this index
loccur[chI] = i;
}
}
string sorted_s = s;
sort(sorted_s.begin(), sorted_s.end());
for (int i = 0; i < len; ++i) {
if (s[i] != sorted_s[i]) {
// Character to replace
int chI = sorted_s[i] - 'a';
// Find the last occurrence
// of this character.
int last_occ = loccur[chI];
// Swap this with the last occurrence
swap(s[i], s[last_occ]);
break;
}
}
return s;
}
// Driver code
int main()
{
string s = "geeks";
cout << findSmallest(s);
return 0;
} Java
// Java implementation of the above approach
import java.util.*;
class GFG{
// Function to return the lexicographically
// smallest String that can be formed by
// swapping at most one character.
// The characters might not necessarily
// be adjacent.
static String findSmallest(char []s)
{
int len = s.length;
// Store last occurrence of every character
int []loccur = new int[26];
// Set -1 as default for every character.
Arrays.fill(loccur, -1);
for (int i = len - 1; i >= 0; --i) {
// Character index to fill
// in the last occurrence array
int chI = s[i] - 'a';
if (loccur[chI] == -1) {
// If this is true then this
// character is being visited
// for the first time from the last
// Thus last occurrence of this
// character is stored in this index
loccur[chI] = i;
}
}
char []sorted_s = s;
Arrays.sort(sorted_s);
for (int i = 0; i < len; ++i) {
if (s[i] != sorted_s[i]) {
// Character to replace
int chI = sorted_s[i] - 'a';
// Find the last occurrence
// of this character.
int last_occ = loccur[chI];
// Swap this with the last occurrence
char temp = s[last_occ];
s[last_occ] = s[i];
s[i] = temp;
break;
}
}
return String.valueOf(s);
}
// Driver code
public static void main(String[] args)
{
String s = "geeks";
System.out.print(findSmallest(s.toCharArray()));
}
}
// This code is contributed by 29AjayKumarPython3
# Python3 implementation of the above approach
# Function to return the lexicographically
# smallest string that can be formed by
# swapping at most one character.
# The characters might not necessarily
# be adjacent.
def findSmallest(s) :
length = len(s);
# Store last occurrence of every character
# Set -1 as default for every character.
loccur = [-1]*26;
for i in range(length - 1, -1, -1) :
# Character index to fill
# in the last occurrence array
chI = ord(s[i]) - ord('a');
if (loccur[chI] == -1) :
# If this is true then this
# character is being visited
# for the first time from the last
# Thus last occurrence of this
# character is stored in this index
loccur[chI] = i;
sorted_s = s;
sorted_s.sort();
for i in range(length) :
if (s[i] != sorted_s[i]) :
# Character to replace
chI = ord(sorted_s[i]) - ord('a');
# Find the last occurrence
# of this character.
last_occ = loccur[chI];
# Swap this with the last occurrence
# swap(s[i], s[last_occ]);
s[i],s[last_occ] = s[last_occ],s[i]
break;
return "".join(s);
# Driver code
if __name__ == "__main__" :
s = "geeks";
print(findSmallest(list(s)));
# This code is contributed by Yash_RC#
// C# implementation of the above approach
using System;
class GFG{
// Function to return the lexicographically
// smallest String that can be formed by
// swapping at most one character.
// The characters might not necessarily
// be adjacent.
static String findSmallest(char []s)
{
int len = s.Length;
// Store last occurrence of every character
int []loccur = new int[26];
// Set -1 as default for every character.
for (int i = 0; i < 26; i++)
loccur[i] = -1;
for (int i = len - 1; i >= 0; --i) {
// char index to fill
// in the last occurrence array
int chI = s[i] - 'a';
if (loccur[chI] == -1) {
// If this is true then this
// character is being visited
// for the first time from the last
// Thus last occurrence of this
// character is stored in this index
loccur[chI] = i;
}
}
char []sorted_s = s;
Array.Sort(sorted_s);
for (int i = 0; i < len; ++i) {
if (s[i] != sorted_s[i]) {
// char to replace
int chI = sorted_s[i] - 'a';
// Find the last occurrence
// of this character.
int last_occ = loccur[chI];
// Swap this with the last occurrence
char temp = s[last_occ];
s[last_occ] = s[i];
s[i] = temp;
break;
}
}
return String.Join("", s);
}
// Driver code
public static void Main(String[] args)
{
String s = "geeks";
Console.Write(findSmallest(s.ToCharArray()));
}
}
// This code is contributed by sapnasingh4991输出：
eegks