[置顶] 【LeetCode】344. Reverse String 解题报告

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Subject

出处：https://leetcode.com/problems/reverse-string/

Write a function that takes a string as input and returns the string reversed.

Example:
Given s = “hello”, return “olleh”.

Explain

该题目的意思就是一个简单的字符串反转输出。

so easy~~

Solution

solution 1

将string转换成char[] 数组。然后for循环从数组末尾开始向前整合。

public static String reverseString(String s) {
        String result = "";
        char[] ch = s.toCharArray();
        for (int i = ch.length - 1; i >= 0; i--) {
            result += ch[i];
        }
        return result;
    }

public static String reverseString2(String s) {
        String result = "";
        for (int i = s.length() - 1; i >= 0; i--) {
            result += s.charAt(i);
        }
        return result;
    }

solution 2

直接使用StringBuffer类的reverse()方法。

public static String reverseString3(String s) {
    return new StringBuffer(s).reverse().toString();
}

solution 3

solution 1的时间复杂度都是o(n).

实际上我们可以for循环”一半”数据长度。将首位字符交换即可。

此时时间复杂度为o(n/2).

public static String reverseString4(String s) {
        char[] ch = s.toCharArray();
        int halfLength = s.length() / 2;
        char temp;
        for (int i = 0; i < halfLength; i++) {
            temp = ch[s.length() - 1 - i];
            ch[s.length() - 1 - i] = ch[i];
            ch[i] = temp;
        }
        return new String(ch);
    }

solution 4

第四种方案采用【异或运算】。

异或预算满足交换律

关于异或运算的性质及应用，可以参考这篇blog：

http://www.cnblogs.com/suoloveyou/archive/2012/04/25/2470292.html

public static String reverseString5(String s) {
        char[] ch = s.toCharArray();
        int start = 0;
        int end = ch.length - 1;
        while (start < end) { ch[start] = (char) (ch[start] ^ ch[end]);
            ch[end] = (char) (ch[start] ^ ch[end]);
            ch[start] = (char) (ch[start] ^ ch[end]);
            start++;
            end--;
        }
        return new String(ch);
    }

该方案的时间复杂度也是o(n/2).

solution 5

通过栈来做。不过有点大材小用的感觉。O(∩_∩)O~

public static String reverseString6(String s) {
        Stack<Character> stack = new Stack<>();
        char[] ch = s.toCharArray();
        String result = "";
        for (int i = 0; i < ch.length; i++) {
            stack.push(ch[i]);
        }
        for (int i = 0; i < ch.length; i++) {
            result += stack.pop();
        }
        return result;
    }

时间复杂度是o(2n).

solution 6

通过递归的方式来做。

public static String reverseString7(String s) {
        int length = s.length();
        if (length <= 1) {
            return s;
        }
        String leftStr = s.substring(0, length / 2);
        String rightStr = s.substring(length / 2, length);
        return reverseString7(rightStr) + reverseString7(leftStr);
    }