Valid Word Abbreviation Leetcode

Posted 2020-08-29 璨璨要好好学习

Given a non-empty string s and an abbreviation abbr, return whether the string matches with the given abbreviation.

A string such as "word" contains only the following valid abbreviations:

["word", "1ord", "w1rd", "wo1d", "wor1", "2rd", "w2d", "wo2", "1o1d", "1or1", "w1r1", "1o2", "2r1", "3d", "w3", "4"]

Notice that only the above abbreviations are valid abbreviations of the string "word". Any other string is not a valid abbreviation of "word".

Note:
Assume s contains only lowercase letters and abbr contains only lowercase letters and digits.

Example 1:

Given s = "internationalization", abbr = "i12iz4n":

Return true.

 

Example 2:

Given s = "apple", abbr = "a2e":

Return false.

 

 

反正每次自己七改八改然后看top solution就会觉得自己逻辑混乱。。。= =

以下是我写的代码，有点难懂。。。

public class Solution {
    public boolean validWordAbbreviation(String word, String abbr) {
        int index1 = 0;
        int index2 = 0;
        char[] a = word.toCharArray();
        char[] b = abbr.toCharArray();
        while (index1 < a.length && index2 < b.length) {
            if (a[index1] != b[index2]) {
                int num = 0;
                if (index2 < b.length && b[index2] - ‘0‘ <= 0 || b[index2] - ‘0‘ > 9) {
                    return false;
                }
                while (index2 < b.length && b[index2] - ‘0‘ >= 0 && b[index2] - ‘0‘ <= 9) {
                    num = num * 10 + b[index2] - ‘0‘;
                    index2++;
                }
                if (num == 0) {
                    return false;
                } else {
                    int count = 0;
                    while (count < num) {
                        if (count < num && index1 > a.length - 1) {
                            return false;
                        }
                        index1++;
                        count++;
                    }
                }
            } else {
                index1++;
                index2++; 
            }
            
        }
        return index1 == a.length && index2  == b.length;
    }
}

看了下top solution重新写的

public class Solution {
    public boolean validWordAbbreviation(String word, String abbr) {
        int index1 = 0;
        int index2 = 0;
        char[] a = word.toCharArray();
        char[] b = abbr.toCharArray();
        while (index1 < a.length && index2 < b.length) {
            if (a[index1] == b[index2]) {
                index1++;
                index2++;
                continue;
            }
            if (b[index2] - ‘0‘ <= 0 || b[index2] - ‘0‘ > 9) {
                System.out.println(1);
                return false;
            }
            int num = 0;
            while (index2 < b.length && b[index2] - ‘0‘ >= 0 && b[index2] - ‘0‘ <= 9) {
                num = num * 10 + b[index2] - ‘0‘;
                index2++;
            }
            index1 = index1 + num;
            if (index1 < a.length && index2 < b.length && a[index1] != b[index2]) {
                return false;
            }
        }
        return index1 == a.length && index2  == b.length;
    }
}

其中

if (index1 < a.length && index2 < b.length && a[index1] != b[index2]) {
                return false;
            }

这里可以省略的。

整体思路就是如果一样，就continue，不一样就看第一个数字是不是0，是0就false。然后把数字取出来，给index1加上，看看最后两个是不是都恰好把数组遍历了一遍。

反正这道题到时候需要回顾的。。。不是这里错就是那里错。