1.一维DP,jump[i]表示到达位置i剩余的最大步数 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.ma...
} 用错误的数组run code一下,看上去答案应该是正确的,不过是414ms超时,由此发现之前的算法果然太暴力。 于是,类似于递归,从前到后遍历,遇到的第一个i+nums[i]将作为新的所求目标位置,并且将result+1,直到所求位置pos到达0或者1为止。结果1ms,但是仍然超时。这到底是什么我也并不知道。。。 代码如下: 1 2 ...
代码 public class Solution { public boolean canJump(int[] nums) { int max = 0, i = 0; for(i = 0; i <= max && i < nums.length; i++){ max = Math.max(max, nums[i] + i); } return i == nums.length; } } Jump Game II Given an array of non-negative integers, you ar...
Given an array of non-negative integers arr, you are initially positioned at start index of the array. When you are at index i, you can jump to i + arr[i] or i - arr[i], check if you can reach to any index with value 0. Notice that you can not jump outside of the array a...
LeetCode 1340. Jump Game V (Java版; Hard) 题目描述 Given an array of integers arr and an integer d. In one step you can jump from index i to index: i + x where: i + x < arr.length and 0 < x <= d. i - x where: i - x >= 0 and 0 < x <= d. ...
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leetcode / solution / 0000-0099 / 0045.Jump Game II / README_EN.md README_EN.md2.78 KB 一键复制编辑原始数据按行查看历史 ylb提交于4年前.feat: add solutions to lc problem: No.0045. Jump Game II 45. Jump Game II 中文文档 Description ...
leetcode / solution / 0000-0099 / 0055.Jump Game / README_EN.md README_EN.md 2.58 KB 一键复制 编辑 原始数据 按行查看 历史 ylb 提交于 4年前 . feat: add solutions to lc problem: No.0055. Jump Game 55. Jump Game Description Solutions Python3 Java C++ Go C# 55. Jump ...
Constraints: 2 <= stones.length <= 2000 0 <= stones[i] <= 231- 1 stones[0] == 0 stonesis sorted in a strictly increasing order. Copyright ©️ 2025 LeetCode All rights reserved 5.8K 116 Case 1Case 2 stones = [0,1,3,5,6,8,12,17] ...
https://leetcode.cn/problems/jump-game-ii/solution/tiao-yue-you-xi-ii-by-leetcode-solution/ 第一题是力扣原题,有O(n)的贪心方法,不用在内循环里找最大值。_牛客网_牛客在手,offer不愁