LeetCode45 Jump Game II 跳跳游戏

问题描述：
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.

Example:

Input: [2,3,1,1,4]
Output: 2

Explanation: 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.
题源：here；完整代码：here
思路：
贪心算法就可以完成该题：我们找到当前位置所能够到达的所有位置中能到达最远位置的位置。有点绕，举个例子：

nums = { 7, 0, 9, 6, 9, 6, 1, 7, 9, 0, 1, 2, 9, 0, 3 }

我们从0开始，所能到达的最远位置是：0+nums[0]=7，我们在0-7中找到能到达最远位置是7+nums[7]=14，因为nums.size()=14，所以2步即可到达终点。
代码实现如下：

class Solution {
public:
    int jump(vector<int>& nums) {
        if (nums.size() == 1) return 0;
        int steps = 0, oldIdx = 0;
        int nextJump = oldIdx + nums[oldIdx];
        if (nextJump >= nums.size() - 1) return steps + 1;

        while(oldIdx < nums.size()){
            int maxJump = 0, newIdx = oldIdx;
            for (int j = oldIdx; j <= oldIdx + nums[oldIdx]; j++){
                int nextJump = j + nums[j];
                if (nextJump >= nums.size() - 1) return steps+2;
                if (j + nums[j] > maxJump){
                    maxJump = j + nums[j];
                    newIdx = j;
                }
            }

            oldIdx = newIdx;
            steps++;
        }
    }
};