[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　　

原文地址：https://www.cnblogs.com/lightwindy/p/8537259.html