• LeetCode 345. Reverse Vowels of a String


    Write a function that takes a string as input and reverse only the vowels of a string.

    Example 1:
    Given s = "hello", return "holle".

    Example 2:
    Given s = "leetcode", return "leotcede".

    Note:
    The vowels does not include the letter "y".

    题意:翻转给定字符串中的元音字母。英语中的元音字母有五个:a, e, i, o, u. 字母y是半元音,题中已经申明元音字母不包括y.

    思路:用两个标志位left和right,分别从数组的两端开始判断当前字母是否为元音,如果两边字母同时都是元音字母,则交换;如果左边字母不是元音,则左边标志位left加1;如果右边字母不是元音,则右边标志位right加1。另:要考虑字母的大小写

    解法一:用String类的contains(CharSequence s)方法,代码如下:

    public String reverseVowels(String s){
            if(s == null || s.length() == 0)
                return s;
            char[] chars = s.toCharArray();
            int left = 0, right = chars.length - 1;
            String str = "aeiou";
            char ch;
            boolean lFlag, rFlag;
            while(left < right){
                lFlag = str.contains(String.valueOf(chars[left]).toLowerCase());
                rFlag = str.contains(String.valueOf(chars[right]).toLowerCase());
                if(!lFlag)
                    left++;
                else if(!rFlag)
                    right--;
                else if(lFlag && rFlag){
                    ch = chars[right];
                    chars[right] = chars[left];
                    chars[left] = ch;
                    left++;
                    right--;
                }
            }
            return String.valueOf(chars);
        }

    解法二:利用自定义的函数判断字母是否为元音

    点评:运行时间比上一个方法更短

    public String reverseVowels(String s){
            if(s == null || s.length() == 0)
                return s;
            char[] chars = s.toCharArray();
            int left = 0, right = chars.length - 1;
            char ch;
            boolean lFlag, rFlag;
            while(left < right){
                lFlag = isVowels(Character.toLowerCase(chars[left]));
                rFlag = isVowels(Character.toLowerCase(chars[right]));
                if(!lFlag)
                    left++;
                else if(!rFlag)
                    right--;
                else if(lFlag && rFlag){
                    ch = chars[right];
                    chars[right] = chars[left];
                    chars[left] = ch;
                    left++;
                    right--;
                }
            }
            return String.valueOf(chars);
        }
        public boolean isVowels(char ch){
            if(ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u')
                return true;
            else 
                return false;
        }
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  • 原文地址:https://www.cnblogs.com/zeroingToOne/p/7819848.html
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