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. Example 1: Input…

55. Jump Game Total Accepted: 95819 Total Submissions: 330538 Difficulty: Medium 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 t…

# -*- coding: utf8 -*-'''__author__ = 'dabay.wang@gmail.com' 55: Jump Gamehttps://leetcode.com/problems/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 represe…

55. Jump Game 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. Determine if you are able to reach the la…

55. Jump Game 第一种方法: 只要找到一个方式可以到达,那当前位置就是可以到达的,所以可以break class Solution { public: bool canJump(vector<int>& nums) { int length = nums.size(); ) return false; vector<bool> dp(length,false); dp[] = true; ;i < length;i++){ ;j >= ;j--){…

一.题目说明 题目55. Jump Game,给定一组非负数,从第1个元素起,nums[i]表示你当前可以跳跃的最大值,计算能否到达最后一个index.难度是Medium. 二.我的解答 非常惭愧,这个题目我做完,提交n次,除了几次边界错,其他就是Time Limit Exceeded,而且优化也无果. 我的代码: class Solution{ public: bool canJump(vector<int>& nums) { vector<bool> dp(nums.s…

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. Example 1: Input…

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 minimum number of jumps…

我一开始认为这是一道DP的题目.其实,可以维护一个maxReach,并对每个元素更新这个maxReach = max(maxReach, i + nums[i]).注意如果 i>maxReach,说明从起点开始能跳到的最远距离不到i, 所以i后面的点也就无法到达了.另外如果 maxReach >= n-1 说明已经可以跳到终点了,之后的点也就不用继续检查了. class Solution(object): def canJump(self, nums): """…

Jump Game 是一道有意思的题目.题意很简单,给你一个数组,数组的每个元素表示你能前进的最大步数,最开始时你在第一个元素所在的位置,之后你可以前进,问能不能到达最后一个元素位置. 比如: A = [2, 3, 1, 1, 4], return true. 一种走法是 0 - 2 - 3 - 4,还有一种走法是 0 - 1 - 4 O(n ^ 2) 解法 一个很显然,几乎不用动脑的解法. 设置一个布尔数组f,f[0] === true 表示 index === 0 这个位置能够到达,模拟每个…