LeetCode 솔루션 분류
[2/8] 45. Jump Game II
본문
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]andi + 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:
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 <= 1040 <= nums[i] <= 1000
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class Solution:
def jump(self, nums: List[int]) -> int:
steps = 0
curr_end = 0
curr_furthest = 0
for i in range(len(nums)-1):
curr_furthest = max(curr_furthest, nums[i] + i)
if curr_furthest >= len(nums)-1:
steps += 1
break
if i == curr_end:
steps += 1
curr_end = curr_furthest
return steps
# dp solution w/ time O(n) - O(n^2)
# dp = [float('inf')]*len(nums)
# dp[0] = 0
# for i in range(len(nums)):
# for j in range(i, min(i+nums[i]+1, len(nums))):
# dp[j] = min(dp[j],dp[i] + 1)
# return dp[-1]
