Can you solve this real interview question? Jump Game II - You are given a 0-indexed array of integers nums of length n. You are initially positioned at nums[0]. Each element nums[i] represents the maximum length of a forward jump from index i. In other
class Solution { 2 public: 3 int jump(int A[], int n) { 4 // Start typing your C/C++ solution below 5 // DO NOT write int main() function 6 if (n<=0) 7 return true; 8 int *flag = new int[n]; 9 int *index = ...
这样遍历的时间其实只有n 次,时间O(n),空间也是O(n).改进地方就是空间改O(1)。 View Code discuss 中有空间O(1)的实现,逻辑是一样的,就贴上来吧。 https://oj.leetcode.com/discuss/422/is-there-better-solution-for-jump-game-ii View Code...
LeetCode 45. Jump Game II 程序员木子 香港浸会大学 数据分析与人工智能硕士在读 来自专栏 · LeetCode Description 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 ...
来自专栏 · LeetCode刷题笔记 原题 题目:45. Jump Game Ⅱ 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 mi...
[LeetCode]45 Jump Game II publicclassSolution{publicintjump(int[]A){// Solution A:// return jump_Gready(A);// Solution B:returnjump_DP(A);}/// Solution B: DP//privateintjump_DP(int[]A){if(A==null||A.length<=1)return0;intglobal=0;intlocal=0;intstep=0;intlastreach=0;for(...
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【LeetCode】45. Jump Game II 解题报告(Python) id: fuxuemingzhu 目录 题目描述 题目大意 解题方法 贪心 日期 题目地址:https://leetcode.com/problems/reach-a-number/description/ 题目描述 Given an array of non-negative integers, you are initially positioned at the first index of the array....
[leetcode] 45. Jump Game II(hard)原题 题意: 是Jump Game的衍生题(题解),题意求跳到最后一格所需最少步数,(默认所测数据永远可以跳到最后一格)。 思路: 利用贪心,遍历数组,记录在每一步可跳跃到的最大区域。 1.当前步数 超过已覆盖范围,则表示需要进行跳跃,同时更新已覆盖区域,可覆盖区域。 2.当前...
Input:nums = [2,3,1,1,4]Output:trueExplanation:Jump 1 step from index 0 to 1, then 3 steps to the last index. Example 2: Input:nums = [3,2,1,0,4]Output:falseExplanation:You will always arrive at index 3 no matter what. Its maximum jump length is 0, which makes it impossible...