[LeetCode] Jump Game

最新推荐文章于 2025-05-22 14:52:23 发布

转载最新推荐文章于 2025-05-22 14:52:23 发布 · 53 阅读

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原文链接：http://www.cnblogs.com/diegodu/p/4315940.html

本文介绍了一个经典的编程问题：判断能否从数组的起始位置通过特定规则跳跃到最后一个元素。使用贪心算法解决此问题，逐步更新可达最远位置，直至判断能否成功到达终点。

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.

思路一：贪心法确定每个点能够达到的最远距离，并在能够达到的最远距离内，继续迭代，计算最远距离，最后判断maxReach 是否到达n-1处。

 i <= maxReach 这个判断条件非常关键

class Solution {
    public:
        bool canJump(int A[], int n) {
            int maxReach = 0;// the most rigth we can reach
    
            for(int i = 0; i <= maxReach && i < n; i++)
            {   
                //cout << maxReach << endl;
                maxReach = max(maxReach, i + A[i]);
            }   

            
            return maxReach >= n-1;

        }   
};