LeetCode403. Frog Jump】的更多相关文章

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…
leetcode403 我们维护青蛙从某个石头上可以跳那些长度的距离即可 用平衡树维护. 总的复杂度O(n^2logn) class Solution { public: bool canCross(vector<int>& stones) { map<int,int>po; int n=stones.size(); map<int,int>dis[1500]; for(int i=0;i<stones.size();i++) po[stones[i]]=…