一年没有管理博客园了,说来实在惭愧。。
最近开始刷LeetCode,之前没刷过,说来也实在惭愧。。。
刚开始按 AC Rates 从简单到难刷,觉得略无聊,就决定按 Add Date 刷,以后也可能看心情随便选题…(⊙o⊙)…今天做了14年 Add 的三个题,其中 Maximum Product Subarray 着实把我折腾了好一会儿,所以想想还是跟大家分享一下我的解法,也或许有更好的方法,期待大家的分享。就把三个题按 Add Date 的先后顺序分享一下吧。
Add Date 2014-03-05
Given an input string, reverse the string word by word.
For example,
Given s = "the sky is blue
",
return "blue is sky the
".
- What constitutes a word?
A sequence of non-space characters constitutes a word. - Could the input string contain leading or trailing spaces?
Yes. However, your reversed string should not contain leading or trailing spaces. - How about multiple spaces between two words?
Reduce them to a single space in the reversed string.
单纯把 "the sky is blue
" reverse 成 "blue is sky the
" 是件很容易的事情,也是比较经典的题目,不过本题说:
1. s 中的开头和结尾有可能有空格,reverse 后的 string 中要去掉;
2. 连续多个空格要变成一个空格。
我的 code 略折腾,用了两个函数:
1. reverseWord 实现最基本的 reverse,首先整个 reverse,变为“eulb si yks eht”,然后每个 word reverse,遍历两边即可。
2. removeSpace 实现检查和删除空格。遍历一遍即可。
复杂度O(n).
附 code,仅供参考。
1 class Solution { 2 public: 3 void reverseWord(string &s) { //反转字符串 4 string::iterator pre = s.begin(); 5 string::iterator post = s.end()-1; 6 char tmp; 7 while(pre < post) { //反转整个字符串 8 tmp = *pre; 9 *pre = *post; 10 *post = tmp; 11 ++pre; 12 --post; 13 } 14 pre = s.begin(); 15 post = pre; 16 while(post != s.end()) { //反转每个单词 17 while(pre != s.end() && *pre == ' ') { 18 ++pre; 19 } 20 if(pre == s.end()) 21 return; 22 post = pre; 23 while(post != s.end() && *post != ' ') { 24 ++post; 25 } 26 --post; 27 if(post != s.end() && pre < post) { 28 string::iterator p1 = pre; 29 string::iterator p2 = post; 30 while(p1 < p2) { 31 tmp = *p1; 32 *p1 = *p2; 33 *p2 = tmp; 34 ++p1; 35 --p2; 36 } 37 } 38 ++post; 39 pre = post; 40 } 41 } 42 43 void removeSpace(string &s) { //检查和删除空格 44 string::iterator pre = s.begin(); 45 string::iterator post = pre; 46 while(pre != s.end() && *pre == ' ') { //删除开头的空格 47 s.erase(s.begin()); 48 pre = s.begin(); 49 } 50 if(pre == s.end()) 51 return; 52 post = pre; 53 while(post != s.end()) { //把连续多个空格变为一个 54 while(pre != s.end() && *pre != ' ') 55 ++pre; 56 if(pre == s.end()) 57 return; 58 post = pre+1; 59 while(post != s.end() && *post == ' ') { 60 s.erase(post); 61 post = pre+1; 62 } 63 ++pre; 64 } 65 if(!s.empty()) { //如果最后有空格则删除 66 pre = s.end()-1; 67 if(*pre == ' ') 68 s.erase(pre); 69 } 70 } 71 72 void reverseWords(string &s) { 73 reverseWord(s); 74 removeSpace(s); 75 } 76 };