未分类

leetcode - 647. Palindromic Substrings

编程题

Given a string, your task is to count how many palindromic substrings in this string.

The substrings with different start indexes or end indexes are counted as different substrings even they consist of same characters.

Example 1:

1
2
3
4
>Input: "abc"
>Output: 3
>Explanation: Three palindromic strings: "a", "b", "c".
>

Example 2:

1
2
3
4
>Input: "aaa"
>Output: 6
>Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa".
>

平淡无奇的解法：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
class Solution {
    public int countSubString(String str){
        int count = 0;
        int i = 0;
        while (i++ < str.length()) {
            count ++;
        }
        for(int j = 2; j <= str.length(); j++){
            for(int x = 0; x+j<=str.length(); x++) {
                if(isPalindrome(str.substring(x,x+j))){
                    count ++;
                }
            }
        }
        return  count;
    }

    //判断字符串是否属于回文
    public boolean isPalindrome(String subStr) {
        for (int i = 0; i < subStr.length(); i++){
            if(subStr.charAt(i) != subStr.charAt(subStr.length() - 1 - i)){
                return false;
            }
        }
        return true;
    }
}

吊炸天的解法：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
/**来自LeetCode代码提交排行榜**/
class Solution {
    public int countSubstrings(String s) {
        int count = 0;
        boolean[][] isPalindrome = new boolean[s.length()][s.length()];
        for(int i = s.length()-1; i >= 0; i--) {
            for(int j = i; j < s.length(); j++) {
                if(i == j || (s.charAt(i) == s.charAt(j) && (isPalindrome[i+1][j-1] || j-i == 1))) {
                    isPalindrome[i][j] = true;
                    count++;
                }
            }            
        }
        return count;
    }
}

647.Palindromic Substrings

程序员面试题精选100题(71)-回文子字符串的数目 - 算法

分享到