LeetCode_Jump Game II

最新推荐文章于 2022-05-23 16:00:10 发布

shengno1

最新推荐文章于 2022-05-23 16:00:10 发布

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分类专栏： Something_Interesting C++ Leetcode xOJ_Coding 文章标签： C++ leetcode 算法题贪心法

本文链接：https://blog.youkuaiyun.com/sheng_ai/article/details/50118877

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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。

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一.题目

Jump Game II

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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.)

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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;
    }
};