LeetCode - Jump Game

Jump Game

2014.2.26 04:21

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.

Solution:

　　If you‘re standing at position i, you can jump at least 0, and at most a[i] steps forward.

　　If you‘re able to reach position i, you must also be able to reach every position before i. Think about why.

　　Thanks to that conclusion, we only need to record the farthest position we can reach. When the array is scanned for one pass, we check if the farthest position we can reach is n. This algorithm is linear and online.

　　Time complexity is O(n). Space complexity is O(1).

Accepted code:

 1 // 2CE, 1TLE, 1AC, simple online algorithm with O(n) time.
 2 class Solution {
 3 public:
 4     bool canJump(int A[], int n) {
 5         if (A == nullptr || n == 0) {
 6             return false;
 7         } else if (n == 1) {
 8             return true;
 9         }
10         
11         int ll, rr;
12         
13         rr = 0;
14         for (ll = 0; ll < n; ++ll) {
15             if (rr < ll) {
16                 // this position is unreachable
17                 // if this position is unreachable, so are all those behind it
18                 return false;
19             } else {
20                 rr = mymax(rr, ll + A[ll]);
21             }
22         }
23         
24         return true;
25     }
26 private:
27     int mymax(const int x, const int y) {
28         return (x > y ? x : y);
29     }
30 };

LeetCode - Jump Game

原文：http://www.cnblogs.com/zhuli19901106/p/3568215.html