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2014
11-18

LeetCode-Valid Palindrome[字符串]

Valid Palindrome

Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.

For example,
"A man, a plan, a canal: Panama" is a palindrome.
"race a car" is not a palindrome.

Note:
Have you consider that the string might be empty? This is a good question to ask during an interview.

For the purpose of this problem, we define empty string as valid palindrome.

标签: Two Pointers String
分析

代码1

// Leet Code, Valid Palindrome
// 时间复杂度O(n),空间复杂度O(1)
class Solution {
public:
    bool isPalindrome(string s) {
        transform(s.begin(), s.end(), s.begin(), ::tolower);
        auto left = s.begin(), right = prev(s.end());
        while (left < right) {
            if (!::isalnum(*left))  ++left;
            else if (!::isalnum(*right)) --right;
            else if (*left != *right) return false;
            else{ left++, right--; }
        }
        return true;
    }
};

Java代码:

class Solution:
    # @param s, a string
    # @return a boolean
    def isPalindrome(self, s):
        low = 0
        high = len(s)-1
        s = s.lower()
        while low < high:
            while s[low].isalnum() is False and low < high: low += 1
            while s[high].isalnum() is False and low < high: high -= 1
            # print(s[low],"  ", s[high])
            if s[low] != s[high]:
                return False
            low += 1
            high -= 1
        return True

  1. 第二个方法挺不错。NewHead代表新的头节点,通过递归找到最后一个节点之后,就把这个节点赋给NewHead,然后一直返回返回,中途这个值是没有变化的,一边返回一边把相应的指针方向颠倒,最后结束时返回新的头节点到主函数。

  2. 给你一组数据吧:29 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 1000。此时的数据量还是很小的,耗时却不短。这种方法确实可以,当然或许还有其他的优化方案,但是优化只能针对某些数据,不太可能在所有情况下都能在可接受的时间内求解出答案。