标签：contain mile pac solution rac for write tween not
Given an input string, reverse the string word by word.
For example,
Given s = "the sky is blue",
return "blue is sky the".
import java.util.Stack; public class Solution { /* * @param s: A string * @return: A string */ public String reverseWords(String s) { // write your code here if(s == null){ return null; } if(s.length() == 0){ return new String(); } Stack<String> stack = new Stack<String>(); int head = 0; int tail = 0; while(head < s.length()){ while(head < s.length() && s.charAt(head) == ‘ ‘){ head++; } tail = head; while(tail < s.length() && s.charAt(tail) != ‘ ‘){ tail ++; } if(head >= s.length() || tail > s.length()){ break; } stack.push(s.substring(head, tail)); head = tail; } String reversed = new String(); while (!stack.isEmpty()){ reversed += stack.pop(); reversed += " "; } if(reversed.length() > 0){ reversed = reversed.substring(0, reversed.length() -1); } return reversed; } }
1. str.substring(int, int); 而不是str.subString(int, int);
2. substring(int beginIndex,int endIndex)从指定的 beginIndex 处开始，直到索引 endIndex - 1 处的字符。因此，该子字符串的长度为 endIndex-beginIndex。 "smiles".substring(1, 5) returns "mile"
3. 在塞完所有单词到stack里后，最后一个个拿出来打印的时候，切记切记循环体for()里面不可以是int i = 0; i < stack.size(); i++!，因为你for{}里面的操作有在pop，那你等于size这个值就一直在变！可以
int wordCount = stack.size(); if(!stack.isEmpty()){ for(int i = 0; i < wordCount - 1; i++){ reversed+= stack.pop(); reversed+= " "; } reversed += stack.pop(); }
或者
while (!stack.isEmpty()){ reversed += stack.pop(); reversed += " "; } if(reversed.length() > 0){ reversed = reversed.substring(0, reversed.length() -1); }
}
4.注意关注输入处理，null, "", " "。在心里跑一下他们
lintcode53 Reverse Words in a String -easy
标签：contain mile pac solution rac for write tween not
原文地址：http://www.cnblogs.com/jasminemzy/p/7504633.html
