lintcode53 Reverse Words in a String -easy

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

 1. 暴力的stack法。

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

2. 巧妙的利用String里面的split方法：

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 || s.length() == 0){
            return "";
        }
        String res = "";
        String[] split = s.split(" ");
        for (int i = split.length - 1; i >= 0; i--){
            if (!split[i].equals("") && !split[i].equals(" ")){
                res += split[i];
                res += " ";
            }
        }
        
        if (res.length() > 0){
            res = res.substring(0, res.length() - 1);
        }

        return res;
    }
}

1. String.split()方法如果用“.”作为分隔的话,必须是如下写法,String.split("\."),这样才能正确的分隔开,不能用String.split(".”); 类似的还有| *。另外如果在一个字符串中有多个分隔符,可以用“|”作为连字符,比如,“acount=? and uu =? or n=?”,把三个都分隔出来,可以用String.split("and|or");
2. 切记String的对比不要用 ==， != 。而是需要用string.equals(“xxx”);