Re: [閒聊] 每日LeetCode已回收

看板Marginalman作者Rushia (みけねこ的鼻屎)時間2年前 (2023/07/03 23:38)推噓3(3推 0噓 2→)

留言5則, 4人參與討論串359/719 (看更多)

https://leetcode.com/problems/buddy-strings/ 859. Buddy Strings 給你一個字串 s 和一個字串 goal，如果 s 的兩個字元交換一次後等於 goal 則他是一個 Buddy Strings，判斷 s 是否是一個 Buddy Strings。 Example 1: Input: s = "ab", goal = "ba" Output: true Explanation: You can swap s[0] = 'a' and s[1] = 'b' to get "ba", which is equal to goal. Example 2: Input: s = "ab", goal = "ab" Output: false Explanation: The only letters you can swap are s[0] = 'a' and s[1] = 'b', which results in "ba" != goal. Example 3: Input: s = "aa", goal = "aa" Output: true Explanation: You can swap s[0] = 'a' and s[1] = 'a' to get "aa", which is equal to goal. 思路： 1.一個一個排掉corner case Java Code： ------------------------------------------------- class Solution { public boolean buddyStrings(String s, String goal) { // 如果兩個字串長度不同怎麼交換都失敗 if (s.length() != goal.length()) { return false; } // 比較s和goal不同的字元數 int notSame = 0; for (int i = 0; i < s.length(); i++) { if (s.charAt(i) != goal.charAt(i)) { notSame++; } } // 不同的數量大於兩個怎麼交換都失敗 if (notSame > 2) { return false; } // 統計字母數 int[] cnt1 = new int[26]; int[] cnt2 = new int[26]; for (int i = 0; i < s.length(); i++) { cnt1[s.charAt(i) - 'a']++; cnt2[goal.charAt(i) - 'a']++; } // 檢查字母數有幾個相同 int same = 0; for (int i = 0; i < 26; i++) { if (cnt1[i] == cnt2[i]) { same++; } } // 如果數量全相同只要有任意字母數量大於2就可以原地交換 if (same == 26) { for (int i = 0; i < 26; i++) { if (cnt1[i] > 1) { return true; } } } // 找到第一個不同的字元，如果s和goal有相同數量的字元則合法 for (int i = 0; i < s.length(); i++) { if (s.charAt(i) != goal.charAt(i)) { return cnt1[s.charAt(i) - 'a'] == cnt2[s.charAt(i) - 'a'] && cnt1[goal.charAt(i) - 'a'] == cnt2[goal.charAt(i) - 'a']; } } return false; } } ------------------------------------------------- 難度：easy 幹你娘機掰漬鯊 == -- https://i.imgur.com/PIoxddO.jpg