leetcode (Reverse Vowels of a String)

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Title:Reverse Vowels of a String   345

Difficulty:Easy

原题leetcode地址:https://leetcode.com/problems/reverse-vowels-of-a-string/

1.    双指针

时间复杂度:O(n),一次一层while循环,最长遍历整个数组。

空间复杂度:O(n),申请StringBuilder和字符数组。

    /**
     * 双指针
     * @param s
     * @return
     */
    public static String reverseVowels(String s) {

        if (s.length() == 0) {
            return "";
        }

        int begin = 0;
        int end = s.length() - 1;
        char sChar[] = s.toCharArray();
        StringBuilder sb = new StringBuilder();

        while (begin < end) {
            if (isVowels(sChar[begin]) && isVowels(sChar[end])) {
                char tmp = sChar[begin];
                sChar[begin] = sChar[end];
                sChar[end] = tmp;
                begin++;
                end--;
            }
            if (!isVowels(sChar[begin]) && isVowels(sChar[end])) {
                begin++;
            }
            if (isVowels(sChar[begin]) && !isVowels(sChar[end])) {
                end--;
            }
            if (!isVowels(sChar[begin]) && !isVowels(sChar[end])) {
                begin++;
                end--;
            }

        }

        for (int i = 0; i < sChar.length; i++) {
            sb.append(sChar[i]);
        }

        return sb.toString();

    }

    private static boolean isVowels(char c) {

        switch (c) {
            case 'a':
            case 'A':
            case 'e':
            case 'E':
            case 'i':
            case 'I':
            case 'o':
            case 'O':
            case 'u':
            case 'U':
                return true;
            default :
                return false;
        }

    }

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转载自blog.csdn.net/hsx1612727380/article/details/85010847