inspired by the discussion in leetcode andhttp://tech-wonderland.net/blog/leetcode-jump-game-ii.html the keypoint of solving the problem by greedy approach is that we should keep the current maxium reachable distance, the next maxium reachable distance and also the steps needed to do it. w...
总结,第一反应的算法不是好算法,要仔细想想,其实这个思路跟jump game那个差不多,只是思路更隐蔽。
Explanation: The minimum number of jumps to reach the last index is 2. Jump 1 step from index 0 to 1, then 3 steps to the last index. Note: You can assume that you can always reach the last index. 感觉这种题目叫做贪心? CPP class Solution { public: int jump(vector<int>& nums) {...
具体代码例如以下: publicclassSolution{publicintjump(int[]nums){/** * 本题用贪心法求解, * 贪心策略是在每一步可走步长内。走最大前进的步数 */if(nums.length<=1){return0;}intindex,max=0;intstep=0,i=0;while(i<nums.length){//假设能直接一步走到最后,直接步数+1结束if(i+nums[i]>=nums...
所属专辑:LeetCode算法题目讲解 喜欢下载分享 声音简介[LeetCode] 45. Jump Game II 跳跃游戏之二博客园:https://www.cnblogs.com/grandyang/p/4373533.htmlGitHub:https://github.com/grandyang/leetcode/issues/45个人网页:https://grandyang.com/leetcode/45/ ...
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] 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
55. 跳跃游戏 - 给你一个非负整数数组 nums ,你最初位于数组的 第一个下标 。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
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 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 position. Your goal is...
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...