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# 45. Jump Game II

## 1.問題

* **需要花幾步才能到達最後一格**

![](/files/-LNp04ewveOK75Y9xHgl)

## 2.想法 <a href="#id-2-xiang-fa" id="id-2-xiang-fa"></a>

* 提問
* function header, parameter
* test input
* 說明想法
  * 當"當前的索引"大於"上一次所能到達的最遠索引時",  將計數器+1, 並且將"上一次所能到達的最遠索引"更新為"當前所能到達的最遠索引"
  * 每走一步都更新"當前所能到達的最遠索引"
* 測試計算複雜度

## **3.程式碼** <a href="#id-3-cheng-shi" id="id-3-cheng-shi"></a>

```
class Solution {
public:
    int jump(vector<int>& nums) {
        int n = nums.size();
        if (n == 0) {
            return 0 ;
        }
        
        int res = 0, last = 0, curr = 0;
        
        for (int i = 0; i < n; i++) {
            if (i > last) {
                last = curr;
                res++;
            }   
            curr = max(curr, i + nums[i]);
        }
        
        return res;
    }
};
```