45. 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.)

题意和55题jump game差不多，但是这里是要求最小的跳数。所以还是可以用贪心算法，但是这里需要对位移进行一些控制，下面有两种做法。

做法一是划定范围[left, right]，在这个范围内，对每个序列的值都要进行一次运算和比较，假如是到达了末尾，那么就返回，不然的话就取最远的right，然后把之前的right+1作为left，继续下去，代码如下。

Code(LeetCode运行13ms):

int jump(vector<int>& nums) {        if (nums.size() == 1) {            return 0;        }        int left = 0;        int right = 0;        int step = 0;                while (left <= right) {            step++;            int old_right = right;            for (int i = left; i <= old_right; i++) {                int new_right = nums[i] + i;                if (new_right >= nums.size() - 1) {                    return step;                }                if (new_right > right) {                    right = new_right;                }            }            left = old_right + 1;        }        return 0;    }

做法二：其实基本思想和做法一差不多，也是更新右边界是有一定的条件，需要i>最后的下标，然后每次都要更新cur，最后取cur的最大值，代码如下。

Code(LeetCode运行12ms):

int jump(vector<int>& nums) {        int step = 0;        int last_index = 0;        int cur_index = 0;        for (int i = 0; i < nums.size(); i++) {            if (i > last_index) {                step++;                last_index = cur_index;            }            cur_index = max (cur_index, nums[i] + i);        }        return step;    }