LeetCode.45 Jump Game II
来源:互联网 发布:mac 网线转接头 驱动 编辑:程序博客网 时间:2024/05/19 18:48
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.
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.)
Note:
You can assume that you can always reach the last index.
第一次的思路是从数组头部开始遍历,直到该点提供的点数能够使我们从该点直接跳到终点(nums[i] + i >= position),则记录步数count++。并重新标记终点为当前点,继续循环,直到终点为数组起始点为止。
代码如下:
int jump(vector<int>& nums){int count = 0, position = nums.size() - 1;while(position > 0){for(int i = 0; i < position; i++){if(i + nums[i] >= position){position = i;count++;break;}}}return count;}
但是上述代码在最后一个测试用例中提示超时了,我把该用例拷贝到本机测试仍然能得到正确的结果,原因应该是这道题对代码复杂度做了限制,上述代码的复杂度为O(n^2),即可能存在复杂度更低的算法。
重新思考过后换了个思路,依旧是从数组头部开始遍历,记第一个点包含的点数为从该点开始所能到达的最远位置(mark),然后开始移动指针i,逐个计算到mark位置之前所有点的最远位置,只取较大值记为endPoint,每当指针i指向mark则当前阶段结束,更新mark值,使其等于endPoint,步数step++,继续下一个阶段。直到mark >= nums.size() - 1为止。
代码如下:
int jump(vector<int>& nums){if(nums.size() == 1) return 0;int step = 0, endPoint = 0, mark = nums[0];for(int i = 1; i <= mark; i++){endPoint = nums[i] + i > endPoint ? nums[i] + i : endPoint;if(mark >= nums.size() - 1){step++;return step;}if(i == mark){mark = endPoint;step++;}}return step;}
- [leetcode 45] Jump Game II
- LeetCode(45) Jump Game II
- [LeetCode 45]Jump Game II
- leetcode || 45、 Jump Game II
- leetcode 45:Jump Game II
- 【LeetCode】45:Jump Game II
- leetcode 45: Jump Game II
- LeetCode-45 Jump Game II
- Leetcode#45||Jump Game II
- leetcode-45 Jump Game II
- Leetcode #45 Jump Game II
- leetcode 45:Jump Game II
- LeetCode 45 - Jump Game II
- 【LeetCode-45】Jump Game II
- LeetCode 45 Jump Game II
- leetcode(45):Jump Game II
- Leetcode 45 Jump Game II
- Leetcode (45) Jump Game II
- nodejs 中读取文件状态;fs.stat()方法;
- redis启动出错Creating Server TCP listening socket *:6379: bind: No error
- linux input子系统的的输入类型
- 关于控件与布局
- 基于蚁群的无线传感器网络路由协议研究
- LeetCode.45 Jump Game II
- 欢迎使用CSDN-markdown编辑器
- MySQL数据恢复和复制对InnoDB锁机制的影响
- CentOS系统yum源使用报错:Error: Cannot retrieve repository metadata (repomd.xml) for repository: rpmforge.
- 商城项目第一天
- 网页端禁止文字复制
- leetcode 153. Find Minimum in Rotated Sorted Array
- ART深入浅出6--了解Dex文件格式(3)
- session与cookies 面试笔记