面试题 58 - I. 翻转单词顺序

输入: "the sky is blue"
输出: "blue is sky the"

输入: "  hello world!  "
输出: "world! hello"
解释: 输入字符串可以在前面或者后面包含多余的空格，但是反转后的字符不能包括。

输入: "a good   example"
输出: "example good a"
解释: 如果两个单词间有多余的空格，将反转后单词间的空格减少到只含一个。

class Solution:
    def reverseWords(self, s: str) -> str:
        if s is None:
            return s
        return ' '.join(list(filter(lambda x: x != '', s.strip(' ').split(' ')))[::-1])

class Solution {
    public String reverseWords(String s) {
        if (s == null || s.length() == 0) {
            return s;
        }
        String[] words = s.split("\\s+");
        StringBuilder sb = new StringBuilder();
        int len = words.length;
        for (int i = len - 1; i >= 0; --i) {
            if (!"".equals(words[i])) {
                sb.append(words[i]).append(" ");
            }
        }
        s = sb.toString();
        len = s.length();
        return len > 0 ? s.substring(0, len - 1) : "";
    }
}

/**
 * @param {string} s
 * @return {string}
 */
var reverseWords = function (s) {
  return s
    .split(" ")
    .reduce((acc, cur) => (cur !== "" ? acc.concat(cur) : acc), [])
    .reverse()
    .join(" ");
};

func reverseWords(s string) string {
	s = strings.Trim(s, " ")
	n := len(s) - 1
	builder := new(strings.Builder)
	for i, j := n, n; i >= 0; j = i {
		for i >= 0 && s[i] != ' ' {
			i--
		}
		if builder.Len() != 0 {
			builder.WriteRune(' ')
		}
		builder.WriteString(s[i+1 : j+1])
		for i >= 0 && s[i] == ' ' {
			i--
		}
	}
	return builder.String()
}