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Sign up【每日一题】- 2020-01-15 - 925. 长按键入 #276
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控制双指针对两个字符串进行字符比对即可。 public boolean isLongPressedName(String name, String typed) {
if (typed.length() < name.length())
return false;
int left = 0, right = 0;
while (left < name.length() && right < typed.length()) {
if (name.charAt(left) == typed.charAt(right)) {
left++;
right++;
} else if (right > 0 && typed.charAt(right) == typed.charAt(right - 1))
right++;
else
return false;
}
return left == name.length();
} |
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@unclegem same idea Python Code class Solution:
def isLongPressedName(self, name: str, typed: str) -> bool:
i = j = 0
m = len(name)
n = len(typed)
if m > n:
return False
while i < m and j < n:
if name[i] == typed[j]:
i += 1
j += 1
elif typed[j] == typed[j - 1]:
j += 1
else:
return False
return i == m |
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public boolean isLongPressedName(String name, String typed) { |
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@hhbszyc001 注意下格式 |
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你的朋友正在使用键盘输入他的名字 name。偶尔，在键入字符 c 时，按键可能会被长按，而字符可能被输入 1 次或多次。
你将会检查键盘输入的字符 typed。如果它对应的可能是你的朋友的名字（其中一些字符可能被长按），那么就返回 True。
示例 1：
输入：name = "alex", typed = "aaleex"
输出：true
解释：'alex' 中的 'a' 和 'e' 被长按。
示例 2：
输入：name = "saeed", typed = "ssaaedd"
输出：false
解释：'e' 一定需要被键入两次，但在 typed 的输出中不是这样。
示例 3：
输入：name = "leelee", typed = "lleeelee"
输出：true
示例 4：
输入：name = "laiden", typed = "laiden"
输出：true
解释：长按名字中的字符并不是必要的。
提示：
name.length <= 1000
typed.length <= 1000
name 和 typed 的字符都是小写字母。
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/long-pressed-name
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。