55. Jump Game

Posted 三颗心

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了55. Jump Game相关的知识,希望对你有一定的参考价值。

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.

For example:
A = [2,3,1,1,4], return true.

A = [3,2,1,0,4], return false.

贪心法,每次跳跃一步,可跳步数减去1,如果新的位置步数大于剩余步数,使用新的步数继续移动,如果可跳步数小于0且还没到最后一个步数,那么失败。

 1 class Solution {
 2 public:
 3     bool canJump(vector<int>& nums) {
 4         if(nums.size() == 0){
 5             return true;
 6         }
 7         
 8         int reach = nums[0];
 9         for(int i = 1; i < nums.size(); i ++){
10             reach--;
11             
12             if(reach < 0){
13                 return false;
14             }
15             
16             if(reach < nums[i]){
17                 reach = nums[i];
18             }
19         }
20         return true;
21     }
22 };

 

以上是关于55. Jump Game的主要内容,如果未能解决你的问题,请参考以下文章

55. 45. Jump Game II *HARD*

[leetcode][55] Jump Game

Leetcode55 Jump Game

Leetcode 55. Jump Game

55. Jump Game

55. Jump Game