二刷55

Author: diguage

docs/0055-jump-game.adoc

@@ -1,37 +1,62 @@
 [#0055-jump-game]
-= 55. Jump Game
+= 55. 跳跃游戏
 
-{leetcode}/problems/jump-game/[LeetCode - Jump Game^]
+https://leetcode.cn/problems/jump-game/[LeetCode - 55. 跳跃游戏 ^]
 
-Given an array of non-negative integers, you are initially positioned at the first index of the array.
+给你一个非负整数数组 `nums` ，你最初位于数组的 *第一个下标*。数组中的每个元素代表你在该位置可以跳跃的最大长度。
 
-Each element in the array represents your maximum jump length at that position.
+判断你是否能够到达最后一个下标，如果可以，返回 `true` ；否则，返回 `false`。
 
-Determine if you are able to reach the last index.
+*示例 1：*
 
-*Example 1:*
+....
+输入：nums = [2,3,1,1,4]
+输出：true
+解释：可以先跳 1 步，从下标 0 到达下标 1, 然后再从下标 1 跳 3 步到达最后一个下标。
+....
 
-[subs="verbatim,quotes,macros"]
-----
-*Input:* [2,3,1,1,4]
-*Output:* true
-*Explanation:* Jump 1 step from index 0 to 1, then 3 steps to the last index.
-----
+*示例 2：*
 
-*Example 2:*
+....
+输入：nums = [3,2,1,0,4]
+输出：false
+解释：无论怎样，总会到达下标为 3 的位置。但该下标的最大跳跃长度是 0 ， 所以永远不可能到达最后一个下标。
+....
 
-[subs="verbatim,quotes,macros"]
-----
-*Input:* [3,2,1,0,4]
-*Output:* false
-*Explanation:* You will always arrive at index 3 no matter what. Its maximum
-             jump length is 0, which makes it impossible to reach the last index.
-----
+*提示：*
+
+* `1 \<= nums.length \<= 10^4^`
+* `0 \<= nums[i] \<= 10^5^`
+
+
+== 思路分析
 
+遍历数组，记录当前位置能到达的最远距离，如果最远距离超过了数组长度，则可以。如果中途最远距离小于等于当前坐标，则无法向前，如果还没到达数组尾部，则是不能到达。
 
 [[src-0055]]
+[tabs]
+====
+一刷::
++
+--
 [{java_src_attr}]
 ----
 include::{sourcedir}/_0055_JumpGame.java[tag=answer]
 ----
+--
+
+二刷::
++
+--
+[{java_src_attr}]
+----
+include::{sourcedir}/_0055_JumpGame_2.java[tag=answer]
+----
+--
+====
+
+
+== 参考资料
 
+. https://leetcode.cn/problems/jump-game/solutions/203549/tiao-yue-you-xi-by-leetcode-solution/[55. 跳跃游戏 - 官方题解^]
+. https://leetcode.cn/problems/jump-game/solutions/46397/pythonji-bai-97kan-bu-dong-ni-chui-wo-by-mo-lan-4/[55. 跳跃游戏 - Python 击败 97%，看不懂你锤我^]

logbook/202503.adoc

@@ -728,6 +728,11 @@ endif::[]
 |{doc_base_url}/0080-remove-duplicates-from-sorted-array-ii.adoc[题解]
 |✅ 双指针，同时需要记录每个元素的次数，相同元素只复制两个。
 
+|{counter:codes2503}
+|{leetcode_base_url}/jump-game/[55. 跳跃游戏^]
+|{doc_base_url}/0055-jump-game.adoc[题解]
+|✅ 贪心算法。没想到这就是贪心？
+
 |===
 
 截止目前，本轮练习一共完成 {codes2503} 道题。

src/main/java/com/diguage/algo/leetcode/_0055_JumpGame.java

@@ -39,6 +39,9 @@ public class _0055_JumpGame {
      * Runtime: 1 ms, faster than 99.35% of Java online submissions for Jump Game.
      *
      * Memory Usage: 37.7 MB, less than 100.00% of Java online submissions for Jump Game.
+     *
+     * @author D瓜哥 · https://www.diguage.com
+     * @since 2019-10-23 12:25
      */
     public boolean canJump(int[] nums) {
         if (Objects.isNull(nums) || nums.length <= 1) {

src/main/java/com/diguage/algo/leetcode/_0055_JumpGame_2.java

@@ -0,0 +1,29 @@
+package com.diguage.algo.leetcode;
+
+public class _0055_JumpGame_2 {
+  // tag::answer[]
+
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2025-05-19 14:50:01
+   */
+  public boolean canJump(int[] nums) {
+    int max = nums[0];
+    for (int i = 0; i < nums.length; i++) {
+      max = Math.max(max, i + nums[i]);
+      if (max >= nums.length - 1) {
+        return true;
+      }
+      if (max < i) {
+        return false;
+      }
+    }
+    return false;
+  }
+
+  // end::answer[]
+  public static void main(String[] args) {
+    new _0055_JumpGame_2()
+      .canJump(new int[]{3, 2, 1, 0, 4});
+  }
+}