题目大意就是,给定一个数组,数组中数字从小到大排列,第一个元素一定是0,青蛙的初始位置就在0,后面依次从小到大排列,表示第几个石子,青蛙只有跳到最后一个石子上才算成功过河,而且青蛙第一次从0位置只能跳一个单位.假如上一次青蛙跳了k个单位,下一次青蛙只能跳 k - 1 , k , k + 1个单位. 这个题的状态不是一个简单的dp数组,因为状态的提取很难用数组实现,这个题要用set和map这种现成的数据结构,这里不再细讲因为这个题只要会C++中的set和map两种存储结构就能简单的AC,map用来…

A frog is crossing a river. The river is divided into x units and at each unit there may or may not exist a stone. The frog can jump on a stone, but it must not jump into the water. Given a list of stones' positions (in units) in sorted ascending ord…

A frog is crossing a river. The river is divided into x units and at each unit there may or may not exist a stone. The frog can jump on a stone, but it must not jump into the water. Given a list of stones' positions (in units) in sorted ascending ord…

A frog is crossing a river. The river is divided into x units and at each unit there may or may not exist a stone. The frog can jump on a stone, but it must not jump into the water. Given a list of stones' positions (in units) in sorted ascending ord…

A frog is crossing a river. The river is divided into x units and at each unit there may or may not exist a stone. The frog can jump on a stone, but it must not jump into the water. Given a list of stones' positions (in units) in sorted ascending ord…

A frog is crossing a river. The river is divided into x units and at each unit there may or may not exist a stone. The frog can jump on a stone, but it must not jump into the water. Given a list of stones' positions (in units) in sorted ascending ord…

A frog is crossing a river. The river is divided into x units and at each unit there may or may not exist a stone. The frog can jump on a stone, but it must not jump into the water. Given a list of stones' positions (in units) in sorted ascending ord…

A frog is crossing a river. The river is divided into x units and at each unit there may or may not exist a stone. The frog can jump on a stone, but it must not jump into the water. Given a list of stones' positions (in units) in sorted ascending ord…

There is a frog staying to the left of the string s=s1s2…sn consisting of n characters (to be more precise, the frog initially stays at the cell 0). Each character of s is either ‘L’ or ‘R’. It means that if the frog is staying at the i-th cell and t…

https://leetcode.com/contest/5/problems/frog-jump/ 这个题目,还是有套路的,之前做过一道题,好像是贪心性质,就是每次可以跳多远,最后问能不能跳到最右边,一会找一下这道题(找到了,就是这个55. Jump Game).然后这道题,差不多是相同的意思,但是每次只能跳3个或者2个,跳了之后还要判断那个位置有没有石头,然后还要记录上一次跳的间隔,然后我就想到了要用map做一个位置到index的映射,主要用于o(1)的查找,可以用unordered_map…