1publicintjump(int[] A) {2//Start typing your Java solution below3//DO NOT write main() function4if(A ==null){5return0;6}7intlen =A.length;8if(len == 0 || len == 1){9return0;10}1112intcur = 0;13intnext = 0;14intret = 0;1516for(inti = 0; i < len; i++){17if...
于是有了下面的代码,小数据集直接就AC了,大数据集却卡住了,极端情况下这个算法的复杂度是O(N2),但是侥幸心理让我还是写了这个解法,但是还是没过,悲催,先记录一下吧。 classSolution { public: intjump(intA[],intn) { // Start typing your C/C++ solution below // DO NOT write int main() function ...
LeetCode45. Jump Game II(C++) 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. ...
【LeetCode】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 minimum number of jumps. ...
leetCode 45.Jump Game II (跳跃游戏) 解题思路和方法,JumpGameIIGivenanarrayofnon-negativeintegers,youareinitiallypositionedatthefirstindexofthearray.Eachelementinthearrayrepresen
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/ ...
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 = [3,2,1,0,4] 输出:false 解释:无论怎样,总会到达下标为 3 的位置。但该下标的最大跳跃长度是 0 , 所以永远不可能到达最后一个下标。 提示...
[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
思路:这题是jump game的扩展。这题有两种解法: 第一种是动态规划,定义数组dp,dp[i]表示到达i处的最小跳跃次数。关键在确定dp[i]的值,我们可以遍历i之前的元素值,找到第一次能跳到或跳过i的下标,然后在其基础上加1即可。代码如下: 1classSolution {2public:3intjump(intA[],intn)4{5vector<int> dp(n...
Leetcode 45 Jump Game II 跳跃游戏II 结论:只有在无法到达更远的位置时,才选择跳跃,选择一个可以达到更远位置的位置,跳到这个位置后,再往前跳! 其他解法: 这题是之前那道上一题的延伸,那题是问能不能到达…