408. Valid Word Abbreviation

O(n) time O(1) space

class Solution {
public:
    bool validWordAbbreviation(string word, string abbr) {
        word += " ", abbr += " "; // padding方便处理
        int m = word.length(), n = abbr.length(), i = 0, j = 0;
        for (int s = 0; i < m && j < n; i += s, ++j) {
            if (isdigit(abbr[j])) {
                if (abbr[j] == '0') return false; // 以0开头的数是非法的
                int x = 0; // 一次性把整个数取出来
                while (isdigit(abbr[j])) {
                    x = x * 10 + abbr[j] - '0';
                    ++j;
                }
                --j; // 注意下标要回退一个
                s = x;
            } else {
                if (word[i] != abbr[j]) return false;
                s = 1;
            }
        }
        return i == m && j == n; // 最后判断是否两个指针同时达到结尾
    }
};

class Solution {
public:
    bool validWordAbbreviation(string word, string abbr) {
        word += " ", abbr += " ";
        int m = word.length(), n = abbr.length(), s = 0, i = 0, j = 0;
        for (; i < m && j < n; ++j) {
            if (isdigit(abbr[j])) {
                if (s == 0 && abbr[j] == '0') return false;
                s = s * 10 + abbr[j] - '0';
            } else {
                i += s;
                s = 0;
                if (i < m && word[i] != abbr[j]) return false;
                ++i;
            }
        }
        return i == m && j == n;
    }
};