為您提供了一個長度為 n 的字串 str。列印字串中每個元素的位置,以便它可以形成回文,否則在螢幕上列印訊息「No palindrome」。
Palindrome 是一個單詞,從反向或向後讀取的字元序列與從正向讀取的字元序列相同,例如 MADAM、racecar。
要尋找序列或單字是回文,我們通常將單字的反向儲存在單獨的字串中並比較兩者,如果它們相同,則給定的單字或序列是回文。但是在這個問題中,我們必須列印排列以形成回文中的單字或序列。
就像,有一個字串str = “tinni” 那麼它可以是intni 或nitin 所以我們必須傳回作為從1 開始的索引和結果的任何一個排列順序可以是2 3 1 4 5 或3 2 1 5 4 兩者中的一個。
上述問題需要像下面給出的範例那樣的解決方案-
Input: string str = “baa” Output: 2 1 3 Input: string str = “tinni” Output: 2 3 1 4 5
void printPalindromePos(string &str) START STEP 1: DECLARE vector<int> pos[MAX] STEP 2: DECLARE AND ASSIGN n WITH LENGTH OF str STEP 3: LOOP FOR i = 0 AND i < n AND i++ pos[str[i]].push_back(i+1) END LOOP STEP 4: SET oddCount = 0 STEP 5: DECLARE oddChar STEP 6: LOOP FOR i=0 AND i<MAX AND i++ IF pos[i].size() % 2 != 0 THEN, INCREMENT oddCount BY 1 SET oddChar AS i END IF END FOR STEP 7: IF oddCount > 1 THEN, PRINT "NO PALINDROME" STEP 8: LOOP FOR i=0 AND i<MAX AND i++ DECRLARE mid = pos[i].size()/2 LOOP FOR j=0 AND j<mid AND j++ PRINT pos[i][j] END LOOP END LOOP STEP 9: IF oddCount > 0 THEN, DECLARE AND SET last = pos[oddChar].size() - 1 PRINT pos[oddChar][last] SET pos[oddChar].pop_back(); END IF STEP 10: LOOP FOR i=MAX-1 AND i>=0 AND i-- DECLARE AND SET count = pos[i].size() LOOP FOR j=count/2 AND j<count AND j++ PRINT pos[i][j] STOP
#include <bits/stdc++.h> using namespace std; // Giving the maximum characters const int MAX = 256; void printPalindromePos(string &str){ //Inserting all positions of characters in the given string. vector<int> pos[MAX]; int n = str.length(); for (int i = 0; i < n; i++) pos[str[i]].push_back(i+1); /* find the number of odd elements.Takes O(n) */ int oddCount = 0; char oddChar; for (int i=0; i<MAX; i++) { if (pos[i].size() % 2 != 0) { oddCount++; oddChar = i; } } /* Palindrome can't contain more than 1 odd characters */ if (oddCount > 1) cout << "NO PALINDROME"; /* Print positions in first half of palindrome */ for (int i=0; i<MAX; i++){ int mid = pos[i].size()/2; for (int j=0; j<mid; j++) cout << pos[i][j] << " "; } // Consider one instance odd character if (oddCount > 0){ int last = pos[oddChar].size() - 1; cout << pos[oddChar][last] << " "; pos[oddChar].pop_back(); } /* Print positions in second half of palindrome */ for (int i=MAX-1; i>=0; i--){ int count = pos[i].size(); for (int j=count/2; j<count; j++) cout << pos[i][j] << " "; } } int main(){ string s = "tinni"; printPalindromePos(s); return 0; }
如果我們執行上面的程序,那麼它將產生以下輸出-
2 3 1 4 5
以上是列印出排列好的字元位置,使其成為回文的C程式的詳細內容。更多資訊請關注PHP中文網其他相關文章!