Jump Game II

一. 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.

Difficulty:Hard

TIME：30MIN

解法（贪心算法）

这道题很有意思，虽然很容易就能看出这道题可以用贪心算法来解，但要用最精简的代码来实现其实并不是那么简单。

为何能用贪心算法来解呢，因为最短的步数肯定是需要我们尽量跳得最远。而跳得最远总能找到下一个跳得最远的格子。

int jump(vector<int>& nums) {    int result = 0;    int fringe = 0, farther = 0;    for (int i = 0; i < nums.size() - 1; i++) { //注意这里长度减一是必要的        farther = max(farther, i + nums[i]);  //不停更新下一步能跳的最远值        //但到达这一步的边界，就说明已经跳过一次了，这个时候肯定也找到了下一步能跳的最远值，然后更新边界信息        if(i == fringe) {            result++;            fringe = farther;        }    }    return result;}

代码的时间复杂度为O(n)。