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题目:Given an input string, reverse the string word by word. For example,Given s = "the sky is blue",return "blue is sky the". 要求: 1)首尾有空格的时候,反转后的string要将空格去掉 2)当string有多个连续空格的时候,只保留一个空格. 代码分析: 对多余空格剔除: 思路分析: 1)从原始s 的最末尾开始扫描,如果遇到空格,用whil…
problem: Given an input string, reverse the string word by word. For example: Given s = "the sky is blue", return "blue is sky the". 问题分析:如何准确的找到每一个需要清除的空格的位置pos,以及每个word对应的pos范围? 解决方法:需要反转一个字符串,使用int string.find_last_not_of(char c, in…
Microsoft.Office.Interop.Word 创建word 转载:http://www.cnblogs.com/chenbg2001/archive/2010/03/14/1685746.html 功能总结或者完善. 一.添加页眉 using System; using System.Collections.Generic; using System.ComponentModel; using System.Data; using System.Linq; using System…
之前用Aspose.Word进行Word转PDF发现'\'这个字符会被转换成'¥'这样的错误,没办法只能换个方法了.下面是Microsoft.Office.Interop.Word转PDF的方法: public bool WordToPDF(string sourcePath, string targetPath) { bool result = false; Microsoft.Office.Interop.Word.Application application = new Microsof…
Word Break 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, given s = "leetcode", dict = ["leet", "code"]. Return t…
 1. Word Break 题目链接 题目要求: 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, given s = "leetcode", dict = ["leet", "code&q…
https://leetcode.com/problems/word-break/ 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"…
一. 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&…
Word Ladder Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that: Only one letter can be changed at a time Each intermediate word must exist in the dictionary For example,…
https://leetcode.com/problems/word-search/ class Solution { public: struct Trie{ Trie *next[]; bool isWord; Trie() { this->isWord = false; for(auto &c: next) c = NULL; } }; void insert(Trie *root, string word) { Trie *p = root; for(auto &c: wor…