C++ 程序查找字典顺序最小的旋转序列设置 2

编写代码以查找圆形数组中的字典最小值，例如对于数组 BCABDADAB，字典最小值为 ABBCABDAD
输入约束：1 < n < 1000
例子：

Input:  GEEKSQUIZ
Output: EEKSQUIZG

Input:  GFG
Output: FGG

Input :  CAPABCQ
Output : ABCQCAP

我们已经讨论了字典最小字符串旋转中的 O(n ² Logn) 解决方案 |设置1。这里我们需要找到最小旋转的起始索引，然后打印旋转。

1) Initially assume 0 to be current min 
   starting index.
2) Loop through i = 1 to n-1.
   a) For each i compare sequence starting 
      at i with current min starting index
   b) If sequence starting at i is lexicographically 
      smaller, update current min starting 
      index.

这是算法的伪代码

function findIndexForSmallestSequence(S, n):
    result = 0
    for i = 1:n-1
        if (sequence beginning at i < 
               sequence beginning at result)
            result = i
        end if
    end for
    return result

这是上述算法的实现。

C++

// C++ program to find lexicographically 
// smallest sequence with rotations. 
#include  
using namespace std; 
  
// Function to compare lexicographically 
// two sequence with different starting 
// indexes. It returns true if sequence 
// beginning with y is lexicographically 
// greater. 
bool compareSeq(char S[], int x, int y, int n) 
{ 
    for (int i = 0; i < n; i++) { 
        if (S[x] < S[y]) 
            return false; 
        else if (S[x] > S[y]) 
            return true; 
        x = (x + 1) % n; 
        y = (y + 1) % n; 
    } 
    return true; 
} 
  
// Function to find starting index 
// of lexicographically smallest sequence 
int smallestSequence(char S[], int n) 
{ 
    int index = 0; 
    for (int i = 1; i < n; i++) 
  
        // if new sequence is smaller 
        if (compareSeq(S, index, i, n)) 
  
            // change index of current min 
            index = i; 
  
    return index; 
} 
  
// Function to print lexicographically 
// smallest sequence 
void printSmallestSequence(char S[], int n) 
{ 
    int starting_index = smallestSequence(S, n); 
    for (int i = 0; i < n; i++) 
        cout << S[(starting_index + i) % n]; 
} 
  
// driver code 
int main() 
{ 
    char S[] = "DCACBCAA"; 
    int n = 8; 
    printSmallestSequence(S, n); 
    return 0; 
}

输出：

AADCACBC

时间复杂度： O(n^2)
辅助空间： O(1)
请参阅有关按字典顺序排列的最小旋转序列的完整文章 |设置2了解更多详情！