Leetcode-403 Frog Jump(青蛙过河) 1#definepb push_back2#define_for(i,a,b) for(int i = (a);i < (b);i ++)3classSolution4{5public:6boolcanCross(vector<int>&stones)7{8intsz =stones.size();9if(sz==2&&stones[1]==1)returntrue;1011vector<set<int>>dp(sz);12dp[0].insert(0);...
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 order, determine if the frog ...
leetcode 403. Frog Jump 403. Frog Jump Hard 47256FavoriteShare 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......
如果距离小于jump-1,我们接着遍历下一块石头;如果dist大于jump+1,说明无法跳到下一块石头,m[key]赋值为false,并返回false;如果在青蛙能跳到的范围中,我们调用递归函数,以新位置i为pos,距离dist为jump,如果返回true了,我们给m[key]赋值为true,并返回true。
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/easy-version-java public class Solution { Map mp; private int[] stones; public boolean canCross(int[] s) {...
Can you solve this real interview question? Frog Jump - A frog is crossing a river. The river is divided into some number of 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.
代码C++ Java JavaScript Golang Cclass Solution { public: bool canCross(vector<int>& stones) { int n = stones.size(); vector<vector<int>> dp(n, vector<int>(n)); dp[0][0] = true; for (int i = 1; i < n; ++i) { if (stones[i] - stones[i - 1] > i) { ...
[0,1,2,3,4,8,9,11]Returnfalse.There is no way to jump to the last stoneasthe gap between the 5th and 6th stone is too large. Solution 一开始以为这是一个 backtracking 的问题,确实能解决,但是时间复杂度太高,在 999999 那个case 里报了 TLE,试了许久,包括 worst case 过滤(某个网友的答案...
class Solution { Map<Integer, Integer> map = new HashMap<>(); Map<String, Boolean> meno = new HashMap<>(); public boolean canCross(int[] stones) { if (stones[1] != 1) { return false; } if (stones.length == 2) { return true; } for (int i = 0; i < stones.length; i...
https://leetcode.com/problems/frog-jump/discuss/88858/easy-version-java https://leetcode.com/problems/frog-jump/discuss/88824/Very-easy-to-understand-JAVA-solution-with-explanations https://leetcode.com/problems/frog-jump/discuss/88937/share-my-non-recursive-c-solution-with-simple-comments...