如何在词典上找到先前的排列？

给定一个单词，在字典上找到它的较小排列。例如，在字典上，较小的排列“ 4321”是“ 4312”，下一个较小的排列“ 4312”是“ 4231”。如果字符串按升序排序，则不存在下一个按字典顺序较小的排列。

我们已经讨论了next_permutation()，该字符串修改字符串，以便它存储按字典顺序较小的置换。 STL还提供std :: prev_permutation。如果函数可以将对象重新排列为字典上较小的排列，则返回“ true”。否则，它返回“ false”。

CPP

// C++ program to demonstrate working of
// prev_permutation()
#include 
using namespace std;
 
// Driver code
int main()
{
    string str = "4321";
    if (prev_permutation(str.begin(), str.end()))
        cout << "Previous permutation is " << str;
    else
        cout << "Previous permutation doesn't exist";
    return 0;
}

输出

Previous permutation is 4312

如何编写我们自己的prev_permutation()?
以下是查找先前排列的步骤：

1. 找到最大索引i，使str [i – 1]> str [i]。
2. 找到最大索引j，使j> = i并且str [j]
3. 交换str [j]和str [i – 1]。
4. 从str [i]开始颠倒子数组。

以下是上述步骤的执行情况–

C++

// C++ program to print all permutations with
// duplicates allowed using prev_permutation()
#include 
using namespace std;
 
// Function to compute the previous permutation
bool prevPermutation(string& str)
{
    // Find index of the last element of the string
    int n = str.length() - 1;
 
    // Find largest index i such that str[i - 1] >
    // str[i]
    int i = n;
    while (i > 0 && str[i - 1] <= str[i])
        i--;
 
    // if string is sorted in ascending order
    // we're at the last permutation
    if (i <= 0)
        return false;
 
    // Note - str[i..n] is sorted in ascending order
 
    // Find rightmost element's index that is less
    // than str[i - 1]
    int j = i - 1;
    while (j + 1 <= n && str[j + 1] < str[i - 1])
        j++;
 
    // Swap character at i-1 with j
    swap(str[i - 1], str[j]);
 
    // Reverse the substring [i..n]
    reverse(str.begin() + i, str.end());
 
    return true;
}
 
// Driver code
int main()
{
    string str = "4321";
    if (prevPermutation(str))
        cout << "Previous permutation is " << str;
    else
        cout << "Previous permutation doesn't exist";
    return 0;
}

Java

// Java program to print all
// permutations with duplicates
// allowed using prev_permutation()
 
class GFG {
 
    // Function to compute
    // the previous permutation
    static boolean prevPermutation(char[] str)
    {
 
        // Find index of the last
        // element of the string
        int n = str.length - 1;
 
        // Find largest index i such
        // that str[i - 1] > str[i]
        int i = n;
        while (i > 0 && str[i - 1] <= str[i]) {
            i--;
        }
 
        // if string is sorted in
        // ascending order we're
        // at the last permutation
        if (i <= 0) {
            return false;
        }
 
        // Note - str[i..n] is sorted
        // in ascending order Find
        // rightmost element's index
        // that is less than str[i - 1]
        int j = i - 1;
        while (j + 1 <= n && str[j + 1] <= str[i - 1]) {
            j++;
        }
 
        // Swap character at i-1 with j
        swap(str, i - 1, j);
 
        // Reverse the substring [i..n]
        StringBuilder sb
            = new StringBuilder(String.valueOf(str));
        sb.reverse();
        str = sb.toString().toCharArray();
 
        return true;
    }
 
    static String swap(char[] ch, int i, int j)
    {
        char temp = ch[i];
        ch[i] = ch[j];
        ch[j] = temp;
        return String.valueOf(ch);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        char[] str = "4321".toCharArray();
        if (prevPermutation(str)) {
            System.out.println("Previous permutation is "
                               + String.valueOf(str));
        }
        else {
            System.out.println("Previous permutation"
                               + "doesn't exist");
        }
    }
}
 
// This code is contributed by 29AjayKumar

Python

# Python 3 program to
# print all permutations with
# duplicates allowed
# using prev_permutation()
 
# Function to compute the
# previous permutation
 
 
def prevPermutation(str):
 
    # Find index of the last element
    # of the string
    n = len(str) - 1
 
    # Find largest index i such that
    # str[i - 1] > str[i]
    i = n
    while (i > 0 and str[i - 1] <= str[i]):
        i -= 1
 
    # if string is sorted in ascending order
    # we're at the last permutation
    if (i <= 0):
        return False
 
    # Note - str[i..n] is sorted in
    # ascending order
 
    # Find rightmost element's index
    # that is less than str[i - 1]
    j = i - 1
    while (j + 1 <= n and
           str[j + 1] <= str[i - 1]):
        j += 1
 
    # Swap character at i-1 with j
    str = list(str)
    temp = str[i - 1]
    str[i - 1] = str[j]
    str[j] = temp
    str = ''.join(str)
 
    # Reverse the substring [i..n]
    str[::-1]
 
    return True, str
 
 
# Driver code
if __name__ == '__main__':
    str = "4321"
 
    b, str = prevPermutation(str)
 
    if (b == True):
        print("Previous permutation is", str)
    else:
        print("Previous permutation doesn't exist")
 
# This code is contributed by
# Sanjit_Prasad

C＃

// C# program to print all
// permutations with duplicates
// allowed using prev_permutation()
using System;
 
class GFG {
 
    // Function to compute
    // the previous permutation
    static Boolean prevPermutation(char[] str)
    {
 
        // Find index of the last
        // element of the string
        int n = str.Length - 1;
 
        // Find largest index i such
        // that str[i - 1] > str[i]
        int i = n;
        while (i > 0 && str[i - 1] <= str[i]) {
            i--;
        }
 
        // if string is sorted in
        // ascending order we're
        // at the last permutation
        if (i <= 0) {
            return false;
        }
 
        // Note - str[i..n] is sorted
        // in ascending order Find
        // rightmost element's index
        // that is less than str[i - 1]
        int j = i - 1;
        while (j + 1 <= n && str[j + 1] <= str[i - 1]) {
            j++;
        }
 
        // Swap character at i-1 with j
        swap(str, i - 1, j);
 
        // Reverse the substring [i..n]
        String sb = String.Join("", str);
        sb = reverse(sb);
        str = sb.ToString().ToCharArray();
 
        return true;
    }
 
    static String swap(char[] ch, int i, int j)
    {
        char temp = ch[i];
        ch[i] = ch[j];
        ch[j] = temp;
        return String.Join("", ch);
    }
    static String reverse(String input)
    {
        char[] temparray = input.ToCharArray();
        int left, right = 0;
        right = temparray.Length - 1;
 
        for (left = 0; left < right; left++, right--) {
            // Swap values of left and right
            char temp = temparray[left];
            temparray[left] = temparray[right];
            temparray[right] = temp;
        }
        return String.Join("", temparray);
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        char[] str = "4321".ToCharArray();
        if (prevPermutation(str)) {
            Console.WriteLine("Previous permutation is "
                              + String.Join("", str));
        }
        else {
            Console.WriteLine("Previous permutation"
                              + "doesn't exist");
        }
    }
}
 
// This code is contributed by Rajput-Ji

输出

Previous permutation is 4312