Runtime: 3 ms, faster than 100.00% of Java online submissions for Jump Game. Memory Usage: 39.2 MB, less than 92.53% of Java online submissions for Jump Game. python版 class Solution: def canJump(self, nums: List[int]) -> bool: reach=0 n=len(nums) for i in range(n): if i>re...
Jump Game II -- LeetCode 原题链接: http://oj.leetcode.com/problems/jump-game-ii/ 这道题是Jump Game的扩展,区别是这道题不仅要看能不能到达终点,而且要求到达...[leetcode]Jump Game 问题叙述性说明: Given an array of non-negative integers, you are initially positioned at the first index...
Your goal is to reach the last index in the minimum number of jumps. For example: Given array A = [2,3,1,1,4] 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.) SOLUTION 1: 参考:http://blog.csdn....
如果数组中没有零,一定能到达最后一个元素,反之,如果有零,那么如果0前面的元素没有一个能直接跳到0后面的元素的话,肯定是不能到达最后的元素的。 classSolution { public: boolcanJump(intA[],intn) { // Start typing your C/C++ solution below // DO NOT write int main() function //如果数组中没...
1//solution 3: one pass.2publicboolean canJump(int[] A) {3//4:424if(A ==null) {5returnfalse;6}78intlen =A.length;910intright =0;11for(inti =0; i < A.length; i++) {12right = Math.max(right, i +A[i]);13if(right == len -1) {14returntrue;15}1617if(i ==right)...
class Solution { public int jump(int[] nums) { int jumps = 0; int currentEnd = 0; int farthest = 0; for(int i = 0;i < nums.length - 1; i++){ farthest = Math.max(farthest,i + nums[i]); if(i == currentEnd){ jumps++; currentEnd = farthest; } } return jumps; } } ...
class Solution { public: bool canJump(vector<int>& nums) { int n=nums.size(); if(n<2) return true; vector<bool> reach(n,false); reach[0]=true; for(int i=0;i<n;++i) { if(reach[i]) { for(int j=1;j<=nums[i]&&i+j<n;j++) ...
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) { int step=0,maxsize=0,end=0; ...
publicclassSolution {publicbooleancanJumpFromPosition(intposition,int[] nums) {if(position == nums.length - 1) {//这个是最低层递归的终止条件returntrue; }intfurthestJump = Math.min(position + nums[position], nums.length - 1);//这步没想到for(intnextPosition = position + 1; nextPosition <...
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...