LeetCode - Jump Game II

Jump Game II

2014.2.26 04:35

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.)

Solution:

　　In this problem, we have to calculate the minimal number of steps to reach the last position. In Jump Game we recorded the furthest position reachable, while in this problem the last farthest position will be recorded as well. The reason for that, is whenever a new boundary is found, you‘ll have to go one step further to reach that boundary. We count the number of steps as we update the furthest position.

　　The algorithm is still one-pass and online.

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

Accepted code:

 1 // 1CE, 3WA, 1AC, online algorithm with O(n) time.
 2 class Solution {
 3 public:
 4     int jump(int A[], int n) {
 5         if (A == nullptr || n <= 0) {
 6             return -1;
 7         } else if (n == 1) {
 8             return 0;
 9         }
10         
11         int last_pos, this_pos;
12         int i;
13         int result;
14         
15         last_pos = 0;
16         this_pos = 0;
17         result = 0;
18         for (i = 0; this_pos < n - 1; ++i) {
19             if (i > this_pos) {
20                 return -1;
21             }
22             if (i + A[i] > this_pos) {
23                 if (i > last_pos) {
24                     last_pos = this_pos;
25                     ++result;
26                 }
27                 this_pos = i + A[i];
28             }
29         }
30         
31         return result + 1;
32     }
33 };

LeetCode - Jump Game II

原文：http://www.cnblogs.com/zhuli19901106/p/3568216.html