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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.)
本题是跳跃问题，利用动态规划求解，每次记录最远到达的距离和步数。时间：16ms。代码如下：
class Solution { public: int jump(vector<int>& nums) { size_t n = nums.size(); if (n == 0 || n == 1) return 0; int maxstep = nums[0], step = 1; if (maxstep >= n - 1) return 1; for (int i = 1; i < n; i++){ if (maxstep == 0) return -1; maxstep--; if (maxstep < nums[i]){ maxstep = nums[i]; step++; } if (maxstep + i >= n - 1) return step; } return -1; } };
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原文地址：http://www.cnblogs.com/Scorpio989/p/4584581.html
