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【LeetCode】151. Reverse Words in a String

时间:2018-11-27 10:48:05      阅读:178      评论:0      收藏:0      [点我收藏+]

Difficulty: Medium

 More:【目录】LeetCode Java实现

Description

Given an input string, reverse the string word by word.

Example:  

Input: "the sky is blue",
Output: "blue is sky the".

Note:

  • A word is defined as a sequence of non-space characters.
  • Input string may contain leading or trailing spaces. However, your reversed string should not contain leading or trailing spaces.
  • You need to reduce multiple spaces between two words to a single space in the reversed string.

Follow up: For C programmers, try to solve it in-place in O(1) space.

Intuition

Use two pointers, startint from the end to the begining of the string to get each word. 

Make use of StringBuilder to append each word.

Solution

    public String reverseWords(String s) {
        if(s==null || s.length()==0)
            return s;
        StringBuilder sb= new StringBuilder();
        int j=s.length();
        for(int i=s.length()-1;i>=0;i--){
            if(s.charAt(i)==‘ ‘)
                j=i;
            else if(i==0 || s.charAt(i-1)==‘ ‘){
                if(sb.length()!=0)
                    sb.append(" ");
                sb.append(s.substring(i,j));
            }
        }
        return sb.toString();
    }

 

Complexity

Time complexity : O(n).

Space complexity :  O(n)

What I‘ve learned

1. There is only one space between two words in the reversed string.

2. Learn how to use two pointers to get each word.

 

 More:【目录】LeetCode Java实现

 

【LeetCode】151. Reverse Words in a String

原文:https://www.cnblogs.com/yongh/p/10024856.html

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