45.Jump Game II
===

###### tags: `Medium`,`Array`,`DP`,`Greedy`

[45. Jump Game II](https://leetcode.com/problems/jump-game-ii/)

### 題目描述

You are given a **0-indexed** array of integers `nums` of length `n`. You are initially positioned at `nums[0]`.

Each element `nums[i]` represents the maximum length of a forward jump from index `i`. In other words, if you are at `nums[i]`, you can jump to any `nums[i + j]` where:

* 0 <= `j` <= `nums[i]` and
* `i` + `j` < `n`

Return *the minimum number of jumps to reach* `nums[n - 1]`. The test cases are generated such that you can reach `nums[n - 1]`.

### 範例

**Example 1:**

Input: nums = [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.

**Example 2:**

Input: nums = [2,3,0,1,4]
Output: 2

**Constraints**:

* 1 <= `nums.length` <= 10^4^
* 0 <= `nums[i]` <= 1000

### 解答
#### Javascript

function jump(nums) {
  let max = 0;
  let jump = 0;
  let temp = 0;
  for (let i = 0; i < nums.length - 1; i++) {
    max = Math.max(max, i + nums[i]);
    if (i === temp) {
      jump++;
      temp = max;
    }
  }

  return jump;
}

> [name=Marsgoat][time= Feb 8, 2023]

#### C++
##### 思路:
1. 計算從左走到右邊最小步數
2. BFS計算步數, 可想成第一輪可到達的點是index 0, 第二輪是第一輪可新到達的點, ..., 當一輪新到達的點達終點即結束

class Solution {
public:
    int jump(vector<int>& nums) {
        if(nums.size()==1) return 0;
        int step = 1, start = 1, end = nums[0];
        while(end < nums.size()-1)
        {
            int next_round_start = end+1;
            int next_round_end = end;
            for(int i = start; i <= end; i++)
                next_round_end = max(next_round_end, i+nums[i]);

            start = next_round_start;
            end = next_round_end;

            step++;
        }
        return step;
    }
};

Time: $O(n)$
Extra Space: $O(1)$

> [name=XD] [time= Feb 8, 2023]

#### Python

class Solution:
    def jump(self, nums: List[int]) -> int:
        if len(nums) == 1: return 0
        jumps, cur_end, cur_far = 0, 0, 0

        for i in range(len(nums) - 1):
            cur_far = max(cur_far, i + nums[i])
            if i == cur_end:
                cur_end = cur_far
                jumps += 1

        return jumps

> [name=Ron Chen][time= Feb 9, 2023]

### Reference

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