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.

把一個字符串中的單詞逆序，單詞字符順序不變。

解法1: New array, 新建一個數組，把字符串以空格拆分成單詞存到數組，在把單詞逆序拷貝進新數組。

解法2: One place，不能新建數組，在原數組的基礎上換位。把字符串的所以字符逆序，然後在把每個單詞的字符逆序。

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);
    }
};

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