LeetCode_Jump Game II

最新推荐文章于 2022-05-23 16:00:10 发布

原创最新推荐文章于 2022-05-23 16:00:10 发布 · 1.4k 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#C++ #leetcode #算法题 #贪心法

Something_Interesting 同时被 3 个专栏收录

174 篇文章

订阅专栏

C++

88 篇文章

订阅专栏

Leetcode

45 篇文章

订阅专栏

本文介绍如何使用贪心法解决JumpGame II问题，包括解题技巧、实现代码和体会，涉及数组操作和最小跳跃步数计算。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

一.题目

Jump Game II

My Submissions

Total Accepted: 53222 Total Submissions: 216499 Difficulty: Hard

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.

Your goal is to reach the last index in the minimum number of jumps.

For example:
Given array A = [2,3,1,1,4]

The minimum number of jumps to reach the last index is 2. (Jump 1 step from index 0 to 1, then 3 steps to the last index.)

Subscribe to see which companies asked this question

Show Tags

Have you met this question in a real interview?

Yes

Discuss

二.解题技巧

这道题可以用贪心法来实现，每次选择下下一跳的位置最远的地方来作为下一跳，这样就可以用最小的步数来到达最后一个位置。

三.实现代码

#include <vector>
#include <algorithm>
#include <climits>

using std::vector;
using std::max_element;

class Solution
{
public:
     int jump(vector<int>& nums)
    {
        const vector<int>::size_type SIZE = nums.size();
        const vector<int>::size_type LastIndex = SIZE - 1;

        if (SIZE < 2)
        {
            return 0;
        }

        //
        for (vector<int>::size_type Index = 0; Index < SIZE; ++Index)
        {
            nums[Index] += Index;
        }

        vector<int>::iterator Now = nums.begin();
        int Result = 1;

        while (Now != nums.end())
        {
            if (*Now >= LastIndex)
            {
                return Result;
            }

            if (Now == nums.begin() + *Now)
            {
                return INT_MAX;
            }

            Now = max_element(Now + 1, nums.begin() + *Now + 1);
            ++Result;
        }

        return Result;
    }
};