You are given a 0-indexed integer arraynumsand an integerk. You are initially standing at index0. In one move, you can jump at mostksteps forward without going outside the boundaries of the array. That is, you can jump from indexito any index in the range[i + 1, min(n - 1, i...
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数组中的每个元素代表你在该位置可以跳跃的最大长度。 判断你是否能够到达最后一个下标,如果可以,返回 true ;否则,返回 false。 示例1: 输入:nums = [2,3,1,1,4] 输出:true 解释:可以先跳 1 步,从下标 0 到达下标 1, 然后再从下标 1 跳 3 步到达最后一个下标。 示例2: 输入:nums = [3,2,1...
leetcode 之Jump Game 题目描述: Given an array of non-negative integers, you are initially positioned at the first index of the array. Each elementinthe array represents your maximum jump length at that position. Determineifyou are able to reach the last index. For example: A= [2,3,1,1...
【LeetCode】55. Jump Game 题目: 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. Determine if ...
https://leetcode.com/problems/jump-game/ 思路1:贪心 O(n) 思路就是贪心。子问题就是判断车在第i个position的时候是否可以到达i+1个position,条件就是当前第i个position所加的油gas[i] + diff(就是到达第i个position时剩下来的油,可以看做前面几站提供的补给) 大于等于cost[i]。
[LeetCode] Jump Game 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. Determine if you are able to reach the last index....
[LeetCode] 55. Jump Game 跳跃游戏博客园:https://www.cnblogs.com/grandyang/p/4371526.htmlGitHub:https://github.com/grandyang/leetcode/issues/55个人网页:https://grandyang.com/leetcode/55/, 视频播放量 117、弹幕量 0、点赞数 0、投硬币枚数 0、收藏人数 0、转
1publicbooleancanJump(int[] A) {2//Start typing your Java solution below3//DO NOT write main() function4intlen =A.length;5//using DP6int[] jump =newint[len];7jump[0] = 0;89for(inti = 1; i < len; i++){10jump[i] = Math.max(jump[i - 1], A[i - 1]) - 1;11if(...
leetcode—jump game 1.题目描述 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. Determine if you are able to reach the last index....