# 55. 跳跃游戏

## 题目描述

```输入: [2,3,1,1,4]

```输入: [3,2,1,0,4]

## 关键点解析

• 建模 (记录和更新当前位置能够到达的最大的索引即可)

## 代码

• 语言支持: Javascript，Python3
```/*
* @lc app=leetcode id=55 lang=javascript
*
* [55] Jump Game
*
* https://leetcode.com/problems/jump-game/description/
*
* algorithms
* Medium (31.38%)
* Total Accepted:    252.4K
* Total Submissions: 797.2K
* Testcase Example:  '[2,3,1,1,4]'
*
* Given an array of non-negative integers, you are initially positioned at the
* first index of the array.
*
* Each element in the array represents your maximum jump length at that
* position.
*
* Determine if you are able to reach the last index.
*
* Example 1:
*
*
* Input: [2,3,1,1,4]
* Output: true
* Explanation: Jump 1 step from index 0 to 1, then 3 steps to the last
* index.
*
*
* Example 2:
*
*
* 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.
*
*
*/
/**
* @param {number[]} nums
* @return {boolean}
*/
var canJump = function(nums) {
let max = 0; // 能够走到的数组下标

for(let i = 0; i < nums.length; i++) {
if (max < i) return false; // 当前这一步都走不到，后面更走不到了
max = Math.max(nums[i] + i, max);
}

return max >= nums.length - 1
};```

Python3 Code:

```class Solution:
def canJump(self, nums: List[int]) -> bool:
"""思路同上"""
_max = 0
_len = len(nums)
for i in range(_len-1):
if _max < i:
return False
_max = max(_max, nums[i] + i)
# 下面这个判断可有可无，但提交的时候数据会好看点
if _max >= _len - 1:
return True
return _max >= _len - 1```

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