此时字符串长度必定为奇数，我们不必做任何调整。若有两个字母频次是奇数，比如ab，我们只需要换一次得到aa。以此类推：abc->cbc对应1次；abcd->dccd对应2次；abcde->deced对应2次。所以我们的最少调整次数为x/2，x为奇数频次字母数量（整数除法）。为了获得某个区间内字母频次奇偶性，我们使用前缀和，由于只需要...
Palindrome Partitioning。 当然,还有一部分问题可能需要一些数学知识去解决,或者是需要一些位运算的技巧去快速解决。总之,我们希望找到时间复杂度低的解决方法。为了达到这个目的,我们可能需要在一个解题方法中融合多种思想,比如在 300. Longest Increasing Subsequence 中同时用到了动态规划和二分查找的方法,将复杂度控制...
20, Valid Palindrome: 判断String是否是对称的,只考虑a-z 0-9,特别的是可以将原String用replaceAll(String regex, String replace)来去除掉特别字符,例如String test2 = test.replaceAll("[^a-zA-Z0-9]", ""); 或是test.replaceAll("\\W", ""); 21, Add Binary: still two pointers, just fyi: cha...
1classSolution {2publicbooleancheckPalindromeFormation(String a, String b) {3returncheck(a, b) ||check(b, a);4}56privatebooleancheck(String a, String b) {7inti = 0;8intj = a.length() - 1;9while(i < j && a.charAt(i) ==b.charAt(j)) {10i++;11j--;12}13returnisPalindrome(...
63 O(1) Check Power of 2.java N/A Java [] 64 Paint Fence.java Easy Java [] 65 Palindrome Permutation II.java Medium Java [] 66 Partition Array by Odd and Even.java N/A Java [] 67 Partition Array.java N/A Java [] 68 Partition List.java Medium Java [] 69 Pascal's ...
N/A Shortest Palindrome.java Hard [KMP, String] Java 119 N/A Convert Sorted Array to Binary Search Tree.java Easy [DFS, Divide and Conquer, Tree] Java 120 N/A Populating Next Right Pointers in Each Node.java Medium [DFS, Divide and Conquer, Tree] Java 121 N/A Space Replacement.java...
Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell one share of the stock multiple times) with the following restrictions: You may not engage in multiple transactions at the same time (ie, you must sell the stock before...
慢速模式的一个案例是当你想要确定一个链表是否为回文(palindrome)时。 下面是满足快速和慢速指针模式的问题: 链表循环简单) 回文链表中等) 环形数组的循环(困难) 4.合并 合并区间是一种处理重叠区间的有效技术。 在很多区间的问题中,你既需要找到重叠的区间,也需要在这些区间重叠时合并它们。该模式的...
and check all next elements: +1 +2 …? and on. seems like it is o(N), each element accessed only once Subham Kuamr April 9, 2020 at 2:32 am Nice way to find element in array. there is also good way to find element in arrayFind max element in array ...
As an added challenge, try to code it using only iterators in C++ or iterators in Java. 【解答】只有二维,那问题就不麻烦,用两个指针,一个 p,一个 q,分别指向第一维的位置和第二维的位置。定一个一个 check 方法,用来保持指针指向的位置是有意义的。 代码语言:javascript 代码运行次数:0 运...