jump-game-vii

java

python

python3

mysql

mssql

oraclesql

csharp

javascript

typescript

bash

swift

kotlin

dart

golang

ruby

scala

rust

racket

erlang

elixir

pythondata

postgresql

cangjie

Algorithms

You are given a 0-indexed binary string <code>s</code> and two integers <code>minJump</code> and <code>maxJump</code>. In the beginning, you are standing at index <code>0</code>, which is equal to <code>&#39;0&#39;</code>. You can move from index <code>i</code> to index <code>j</code> if the following conditions are fulfilled:

<ul>
	<li><code>i + minJump &lt;= j &lt;= min(i + maxJump, s.length - 1)</code>, and</li>
	<li><code>s[j] == &#39;0&#39;</code>.</li>
</ul>

Return <code>true</code> if you can reach index <code>s.length - 1</code> in <code>s</code>, or <code>false</code> otherwise.

&nbsp;
Example 1:

<pre>
Input: s = &quot;011010&quot;, minJump = 2, maxJump = 3
Output: true
Explanation:
In the first step, move from index 0 to index 3. 
In the second step, move from index 3 to index 5.
</pre>

Example 2:

<pre>
Input: s = &quot;01101110&quot;, minJump = 2, maxJump = 3
Output: false
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>2 &lt;= s.length &lt;= 105</code></li>
	<li><code>s[i]</code> is either <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
	<li><code>s[0] == &#39;0&#39;</code></li>
	<li><code>1 &lt;= minJump &lt;= maxJump &lt; s.length</code></li>
</ul>

canReach

Jump Game VII

Jump Game II

jump-game-ii

跳跃游戏 II

Jump Game

jump-game

跳跃游戏

Jump Game III

jump-game-iii

跳跃游戏 III

Jump Game IV

jump-game-iv

跳跃游戏 IV

Jump Game V

jump-game-v

跳跃游戏 V

Jump Game VI

jump-game-vi

跳跃游戏 VI

跳跃游戏 VII

Jump Game VIII

jump-game-viii

跳跃游戏 VIII

Count Vowel Strings in Ranges

count-vowel-strings-in-ranges

统计范围内的元音字符串数

Maximum Number of Jumps to Reach the Last Index

maximum-number-of-jumps-to-reach-the-last-index

达到末尾下标所需的最大跳跃次数

String

Dynamic Programming

Prefix Sum

Sliding Window

给你一个下标从 0 开始的二进制字符串 <code>s</code> 和两个整数 <code>minJump</code> 和 <code>maxJump</code> 。一开始，你在下标 <code>0</code> 处，且该位置的值一定为 <code>'0'</code> 。当同时满足如下条件时，你可以从下标 <code>i</code> 移动到下标 <code>j</code> 处：

<ul>
	<li><code>i + minJump <= j <= min(i + maxJump, s.length - 1)</code> 且</li>
	<li><code>s[j] == '0'</code>.</li>
</ul>

如果你可以到达 <code>s</code> 的下标 <code>s.length - 1</code> 处，请你返回 <code>true</code> ，否则返回 <code>false</code> 。

 

示例 1：

<pre>
输入：s = "011010", minJump = 2, maxJump = 3
输出：true
解释：
第一步，从下标 0 移动到下标 3 。
第二步，从下标 3 移动到下标 5 。
</pre>

示例 2：

<pre>
输入：s = "01101110", minJump = 2, maxJump = 3
输出：false
</pre>

 

提示：

<ul>
	<li><code>2 <= s.length <= 105</code></li>
	<li><code>s[i]</code> 要么是 <code>'0'</code> ，要么是 <code>'1'</code></li>
	<li><code>s[0] == '0'</code></li>
	<li><code>1 <= minJump <= maxJump < s.length</code></li>
</ul>