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.
 

Note:

You can assume that you can always reach the last index.

使用dp思想，dfs記憶搜尋，記憶從某個點到最後一個點的最少步數，結果超時。
使用貪心思想，bfs搜尋，不固定目標點，而是每次動態更新陣列，結構正確。

class Solution {
    vector<int> d;
    vector<int> arr;
    int n;

    int dfs(int start) {
        if(start==n-1) return 0;
        if(
 
d[start]) 
            return d[start];
        int res = (1<<30);
        for(int i=start+1;i<n&&i<=start+arr[start];i++)
            res = min(res, 1+dfs(i));
        d[start]=res;
        return res;
    }
    
    int bfs(int start, int end) {
        int last=1;
        for(int
 
 i=start;i<end;i++) {
            while(last<end&&last<=i+arr[i])
                d[last++] = d[i]+1;
        }
        return d[end-1];
    }

    
public:
    int jump(vector<int>& nums) {
        arr = nums;
        n = arr.size();
        d = vector<int>(n, 0);
        // dfs(0); // 超時
        return bfs(0, n);
    }
};