Leetcode题解

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.
For example:
A = [2,3,1,1,4], return true.
A = [3,2,1,0,4], return false.

链接

超时的方法

利用队列存储每个位置某个搜到的所有路线，进行BFS，当到达最后一个元素时返回true。
具体超时的例子：

Last executed input:
[8,2,4,4,4,9,5,2,5,8,8,0,8,6,9,1,1,6,3,5,1,2,6,6,0,4,8,6,0,3,2,8,7,6,5,1,7,0,3,4,8,3,5,9,0,4,0,1,0,5,9,2,0,7,0,2,1,0,8,2,5,1,2,3,9,7,4,7,0,0,1,8,5,6,7,5,1,9,9,3,5,0,7,5]

class Solution {public:    bool canJump(vector<int>& nums) {        queue<int> q;        int sz=nums.size();        vector<int> sum(sz);        for(int i=0;i<sz;i++){            sum[i]=i+nums[i];        }        q.push(0);        while(!q.empty()){            int tmp=q.front();            if(tmp==sz-1) return true;            q.pop();            for(int j=tmp+1;j<=min(sum[tmp],sz-1);j++){                q.push(j);            }        }        return false;         }};

正解

从第一个元素开始记录每个元素能够到达的最远位置，当比较完每个元素之后，在判断能否到达最后一个位置。

class Solution {public:    bool canJump(vector<int>& nums) {        int reach;        int i=0;        for(reach=0;i<nums.size()&& i<=reach;i++){            reach=max(nums[i]+i,reach);        }        return reach>=nums.size()-1;    }};