leetcode-925-easy

Long Pressed Name
思路一: 暴力，遍历两个字符串，对比。边界情况不太好处理

public boolean isLongPressedName(String name, String typed) {
    if (typed.length() < name.length()) return false;

    int left = 0;
    int right = 0;

    while (left < name.length() && right < typed.length()) {
        char t1 = name.charAt(left);
        int count = 0;
        while (left < name.length() && name.charAt(left) == t1) {
            left++;
            count++;
        }

        char t2 = typed.charAt(right);
        int count2 = 0;
        while (right < typed.length() && typed.charAt(right) == t2) {
            right++;
            count2++;
        }

        if (t1 != t2 || count > count2) {
            return false;
        }
    }

    return left == name.length() && right == typed.length();
}