1定义

名称	意义
i	石块在数组stones中的下标
position[i]	第i个石块的position，即stones[i]
max_last_step[i]	到达第i个石块的最后一步所有可能的步长的最大值

2初始状态分析

由于position =1的下一跳position只能是2,3, 因此若可达，两者必有一个有石块，因此初始状态为以下三个状态之一：

(1) 若position = 2, position = 3都有石块，则状态为：

i	1	2	3
position[i]	1	2	3
max_last_step[i]	1	1	2

(2) 若position = 2有石块，position =3没有石块，则状态为：

i	1	2
position[i]	1	2
max_last_step[i]	1	1

(3) 若position = 2没有石块，position =3有石块，则状态为：

i	1	2
position[i]	1	3
max_last_step[i]	1	2

3三个推论

4算法描述

广度优先遍历每个石块。

使用一个二维数组visited记录遍历到的每个石块在数组stones中的下标和上一跳的步长。由推论一，这个二维数组的每个维度的长度都不超过stones数组的长度。

5优化

由推论三，若最后一个石块可达，则stones数组必须满足：

因此，可以先对stones数组遍历一遍，判断该必要条件是否满足。

6算法实现

具体实现中，为了减少减法运算，把二维数组visited中的上一跳步长改为了下一跳最小步长，即上一跳步长-1

#include <stdbool.h>
#include <stdlib.h>struct record {int stone_index;int min_step;
};bool binary_search(int *arr, int size, int elem, int *index)
{int start = 2;int end = size - 1;int mid;while (start <= end){mid = (start + end) >> 1;if (elem == arr[mid]){*index = mid;return true;}if (elem < arr[mid]) end = mid - 1;else start = mid + 1;}*index = start;return false;
}bool can_cross(int* stones, int stonesSize, int **visited, struct record *queue)
{int last_stone = stones[stonesSize - 1];struct record *curr;int front = 0;int tail = 0;int stone_position;int stone_index;bool is_stone;int curr_min_step;curr = &queue[tail++];curr->stone_index = 1;curr->min_step = 0;visited[1][0] = 1;while (front != tail){curr = &queue[front++];curr_min_step = curr->min_step;stone_position = stones[curr->stone_index] + curr_min_step;curr_min_step--;if (stone_position == last_stone || stone_position + 1 == last_stone ||stone_position + 2 == last_stone) return true;is_stone = binary_search(stones, stonesSize, stone_position, &stone_index);if(is_stone && curr_min_step >= 0 && !visited[stone_index][curr_min_step]){curr = &queue[tail++];curr->stone_index = stone_index;curr->min_step = curr_min_step;visited[stone_index][curr_min_step] = 1;}stone_position++;curr_min_step++;if (stone_index < stonesSize && stones[stone_index] < stone_position){stone_index++;}if (stone_index < stonesSize && stones[stone_index] == stone_position && !visited[stone_index][curr_min_step]){curr = &queue[tail++];curr->stone_index = stone_index;curr->min_step = curr_min_step;visited[stone_index][curr_min_step] = 1;}stone_position++;curr_min_step++;if (stone_index < stonesSize && stones[stone_index] < stone_position){stone_index++;}if (stone_index < stonesSize && stones[stone_index] == stone_position && !visited[stone_index][curr_min_step]){curr = &queue[tail++];curr->stone_index = stone_index;curr->min_step = curr_min_step;visited[stone_index][curr_min_step] = 1;}}return false;
}bool canCross(int* stones, int stonesSize) {int **visited;struct record *queue;bool res;int i;if (stones[1] != 1) return false;if (stonesSize == 2) return true;visited = (int **)malloc(sizeof(*visited) * stonesSize);for (i = 0; i < stonesSize; i++){visited[i] = (int *)malloc(sizeof(*visited[0]) * stonesSize);memset(visited[i], 0, sizeof(*visited[0]) * stonesSize);}queue = (struct record *)malloc(sizeof(*queue) * stonesSize * stonesSize);res = can_cross(stones, stonesSize, visited, queue);for (i = 0; i < stonesSize; i++){free(visited[i]);}free(visited);free(queue);return res;
}

403. Frog Jump相关推荐

LeetCode 403 Frog Jump 青蛙过河 Javascirpt 解决方案
代码 /*** @param {number[]} stones* @return {boolean}*/ var canCross = function(stones) {let len = sto ...
Leetcode-403.Frog Jump(青蛙跳石头)
今天的题目是:Leetcode 403-青蛙跳 A frog is crossing a river. The river is divided into x units and at each un ...
算法练习（21）：Frog Jump
题意:给出石头的位置,然后青蛙每一跳的距离a和上一跳距离k满足(k==a||k==a-1||k==a+1),第一个石头是位置0,第一跳只能距离为1. 分析与思路:这个题目用遍历搜索所有的路线是肯定可以 ...
html5 frog jump,frog-jump
https://leetcode.com/problems/frog-jump///受以下网页启发,用递归可行// https://discuss.leetcode.com/topic/59337/e ...
LeetCode All in One 题目讲解汇总(持续更新中...)
原文地址:https://www.cnblogs.com/grandyang/p/4606334.html 终于将LeetCode的大部分题刷完了,真是漫长的第一遍啊,估计很多题都忘的差不多了,这次开 ...
taoqick 搜索自己CSDN博客
L1 L2正则化和优化器的weight_decay参数 kaiming初始化的推导 Pytorch动态计算图 Pytorch自动微分机制 PyTorch中在反向传播前为什么要手动将梯度清零? 通俗讲解 ...
考研词汇用语言记忆
根据目前的时间安排及复习进度,相信大家都泡在英语堆里狂记单词,我也一样不过在偶尔看了下面的单词复习方法后,我试着并为之坚持下来,感觉收获真的不一样因此奉献给大家,希望对大家有所帮助! 告 ...
7000词汇这么背我比较可以接受，连续看20天足以
16天记住7000考研词汇(第一天) 1.With my own ears I clearly heard the heart beat of the nuclear bomb. 我亲耳清楚地听到原子 ...
800个有趣句子帮你记忆7000个单词
800个有趣句子帮你记忆7000个单词 1. With my ownears I clearly heard the heart beat of the nuclear bomb. 我亲耳清楚地听到原 ...
[转]800个有趣句子帮你记忆7000个单词
1. With my own ears I clearly heard the heart beat of the nuclear bomb. 我亲耳清楚地听到原子弹的心脏的跳动. 2. Next y ...

403. Frog Jump

1定义

2初始状态分析

3三个推论

4算法描述

5优化

6算法实现

403. Frog Jump相关推荐

最新文章

热门文章