Given a string, you need to reverse the order of characters in each word within a sentence while still preserving whitespace and initial word order.

Example 1:

Input: "Let's take LeetCode contest"
Output: "s'teL ekat edoCteeL tsetnoc"

Note: In the string, each word is separated by single space and there will not be any extra space in the string.

方法1：(這種比較直接的做法，可能效率會有點低)

class Solution {
    public String reverseWords(String s) {
        StringBuffer sb=new StringBuffer(s);
        StringBuffer newSb=new StringBuffer();
        String [] strs=sb.reverse().toString().split(" +");

        for(int i=strs.length-1;i>=0;i--){
            newSb.append(strs[i]+" ");
         }
         return newSb.toString().trim();       
    }
}
 

時間複雜度：O(n)

空間複雜度：O(n)

方法2：（在方法1的基礎上有一些改進，不過還是利用方法的原理進行計算）

package leetcode;

import org.junit.Test;

/**
 * @author zhangyu
 * @version V1.0
 * @ClassName: ReverseWords
 * @Description: TOTO
 * @date 2018/11/30 14:30
 **/


public class ReverseWords {
    @Test
    public void fun() {
        String s = "we are from china";
        String newString = reverseString(s);
        System.out.println(newString);
    }

    private String reverseString(String s) {
        String[] strs = s.split(" +");
        StringBuilder sb = new StringBuilder();
        for (String ss : strs) {
            sb.append(new StringBuilder(ss).reverse()+" ");
        }
        return sb.toString().trim();
    }
}
 

時間複雜度：O(n)

空間複雜度：O(n)