【leetcode】Word Break (middle)】的更多相关文章

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words. For example, givens = "leetcode",dict = ["leet", "code"]. Return true because &…
思路是这种.我们从第一个字符開始向后依次找,直到找到一个断句的地方,使得当前获得的子串在dict中,若找到最后都没找到.那么就是False了. 在找到第一个后,接下来找下一个断句处,当然是从第一个断句处的下一个字符開始找连续的子串,可是这时与第一个就稍有不同.比方说word='ab', dict={ 'a', ab', ...},在找到a后,接下来处理的是b.我们发现b不在dict中,可是我们发现b能够和a结合,形成ab,而ab在dict中.所以这里的每一个子串就能够有三种选择.要么自己单独作为…
Reverse digits of an integer. Example1: x = 123, return 321Example2: x = -123, return -321 总结:处理整数溢出的方法 ①用数据类型转换long  或 long long ②在每次循环时先保存下数字变化之前的值,处理后单步恢复看是否相等 (比③好) ③整体恢复,看数字是否相等. 思路:注意30000这样以0结尾的数字,注意越界时返回0. 我检查越界是通过把翻转的数字再翻转回去,看是否相等. int rever…
Given a singly linked list L: L0→L1→…→Ln-1→Ln,reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→… You must do this in-place without altering the nodes' values. For example,Given {1,2,3,4}, reorder it to {1,4,2,3}. 思路: 先把链表分成两节,后半部分翻转,然后前后交叉连接. 大神的代码比我的简洁,注意分两节时用快…
Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order. For example,Given the following matrix: [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ] You should return [1,2,3,6,9,8,7,4,5]. 思路:就按照螺旋矩阵的规律 用n记录旋转圈数 每…
Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers. If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order). The replaceme…
There are a total of n courses you have to take, labeled from 0 to n - 1. Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1] Given the total number of courses and a l…
Given a list, rotate the list to the right by k places, where k is non-negative. For example:Given 1->2->3->4->5->NULL and k = 2,return 4->5->1->2->3->NULL. 题目关键点:k可能比链表长度还大, 需要获取链表长度len, 用 k % len 获取具体需要移动的数字. class Solution…
Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x. You should preserve the original relative order of the nodes in each of the two partitions. For example,Given 1->4->3->2…
You are given an n x n 2D matrix representing an image. Rotate the image by 90 degrees (clockwise). Follow up:Could you do this in-place? 思路:我的思路,先沿对角线对称,再左右对称. void rotate(int **matrix, int n) { //先沿对角线对称 ; i < n; i++) { ; j < i; j++) { int tmp = m…