[LeetCode] 151. Reverse Words in a String 翻转字符串中的单词

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

For example,
Given s = "the sky is blue",
return "blue is sky the".

Update (2015-02-12):
For C programmers: Try to solve it in-place in O(1) space.

Clarification:

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.

把一个字符串中的单词逆序，单词字符顺序不变。

解法1: New array, 新建一个数组，把字符串以空格拆分成单词存到数组，在把单词逆序拷贝进新数组。

解法2: One place，不能新建数组，在原数组的基础上换位。把字符串的所以字符逆序，然后在把每个单词的字符逆序。

Java: New array, two pass

public String reverseWords(String s) {  
    String[] words = s.trim().split("\\s+");  
    if(words.length == 0) {  
        return "";  
    }  
    StringBuilder sb = new StringBuilder(words[words.length-1]);  
    for(int i=words.length-2; i >=0; i--) {  
        sb.append(" "+words[i]);  
    }  
    return sb.toString();  
}

Java: New array, one pass

public String reverseWords(String s) {  
    StringBuilder sb = new StringBuilder();  
    int end = s.length();  
    int i = end-1;  
    while(i>=0) {  
        if(s.charAt(i) == ' ') {  
            if(i < end-1) {  
                sb.append(s.substring(i+1, end)).append(" ");  
            }  
            end = i;  
        }  
        i--;  
    }  
    sb.append(s.substring(i+1, end));  
    return sb.toString().trim();  
}

Java: New array, one pass

public String reverseWords(String s) {  
    StringBuilder sb = new StringBuilder();  
    int last = s.length();  
    for(int i=s.length()-1; i>=-1; i--) {  
        if(i==-1 || s.charAt(i)==' ') {  
            String word = s.substring(i+1, last);  
            if(!word.isEmpty()) {  
                if(sb.length() != 0) sb.append(' ');  
                sb.append(word);  
            }  
            last = i;  
        }  
    }  
    return sb.toString();  
}

Java：One place

public String reverseWords(String s) {  
    if(s == null || s.isEmpty()) return s;  
    char[] data = s.toChartArray();  
    int n = data.length;  
    reverse(data, 0, n-1);  
      
    int last = -1;  
    for(int i=0; i<=n; i++) {  
        if(i == n || data[i] == ' ') {  
            if(i-last>1) reverse(data, last+1, i-1);  
            last = i;  
        }  
    }  
      
    return new String(data);  
}  
  
private void reverse(char[] data, int start, int end) {  
    while(start < end) {  
        char tmp = data[start];  
        data[start++] = data[end];  
        data[end--] = tmp;  
    }  
}

Python: New array

class Solution:
    # @param s, a string
    # @return a string
    def reverseWords(self, s):
        return ' '.join(reversed(s.split()))

C++:

class Solution {
public:
    void reverseWords(string &s) {
        int storeIndex = 0, n = s.size();
        reverse(s.begin(), s.end());
        for (int i = 0; i < n; ++i) {
            if (s[i] != ' ') {
                if (storeIndex != 0) s[storeIndex++] = ' ';
                int j = i;
                while (j < n && s[j] != ' ') s[storeIndex++] = s[j++];
                reverse(s.begin() + storeIndex - (j - i), s.begin() + storeIndex);
                i = j;
            }
        }
        s.resize(storeIndex);
    }
};

类似题目：

[LeetCode] 186. Reverse Words in a String II 翻转字符串中的单词 II

[LeetCode] 557. Reverse Words in a String III 翻转字符串中的单词 III　　