Problem

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].

Example 1:

1
2
3
4
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:

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

Constraints:

  • 1 <= nums.length <= 10^4
  • 0 <= nums[i] <= 1000
  • It’s guaranteed that you can reach nums[n - 1].

Solve

和Jump Game I 不同，沒有到不了的問題

法1

想法蠻簡單的做一個DP，紀錄到達這個節點，最少步數

不需要回頭做

但題目要求要為Greedy algo

1
2
3
4
5
6
7
8
9
10
11
12
13
14
class Solution:
    def jump(self, nums: List[int]) -> int:

        dp = {}
        dp[0] = 0

        for i in range(0, len(nums) ):
            
            for j in range(0 , nums[i]+1 ):
                if i+j in dp:
                    continue
                dp[i+j] = dp[i]+1

        return dp[len(nums)-1]

法3 Greedy algo

其實跟法1想法一樣，但更快且Space complexity 更少

不斷更新reach 最遠可以到達的，當到達紀錄的last點時，

再跳躍一次，更新最遠距離，並將count += 1

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
# 網路的解
class Solution:
    def jump(self, nums):
        # Initialize reach (maximum reachable index), count (number of jumps), and last (rightmost index reached)
        reach, count, last = 0, 0, 0
        
        # Loop through the array excluding the last element
        for i in range(len(nums)-1):    
            # Update reach to the maximum between reach and i + nums[i]
            reach = max(reach, i + nums[i])
        
            # If i has reached the last index that can be reached with the current number of jumps
            if i == last:
                # Update last to the new maximum reachable index
                last = reach
                # Increment the number of jumps made so far
                count += 1
        
        # Return the minimum number of jumps required
        return count