Your friend is typing hisname into a keyboard. Sometimes, when typing a characterc, the key might getlong pressed, and the character will be typed 1 or more times.

You examine thetyped characters of the keyboard. ReturnTrueif it is possible that it was your friends name, with some characters (possibly none) being long pressed.

这道题千万要注意一点，不是简单的统计字符的个数，因为这里面还有顺序的因素，比如abc和bca虽然字符相同，但是不符合条件。正确的做法就是two pointers，考虑顺序因素

class Solution {
    public boolean isLongPressedName(String name, String typed) {
        if(name == null && typed == null)
            return true;
        if(name == null || typed == null)
            return true;

        if(typed.length() < name.length())
            return false;

        int i = 0;
        int j = 0;

        while(i < name.length()) {
            char c = name.charAt(i);
            int count1 = 0;
            int count2 = 0;

            while(i < name.length() && name.charAt(i) == c) {
                i++;
                count1++;
            }

            while(j < typed.length() && typed.charAt(j) == c) {
                j++;
                count2++;
            }

            if(count2 < count1)
                return false;
        }

        return true;
    }
}