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55. Jump Game

时间:2016-09-15 14:55:01      阅读:142      评论:0      收藏:0      [点我收藏+]

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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 };

 

55. Jump Game

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原文地址:http://www.cnblogs.com/sankexin/p/5874713.html

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