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[mintheon] Week 09 #990
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[mintheon] Week 09 #990
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class Solution { | ||
// 시간복잡도: O(n) | ||
// 공간복잡도: O(1) | ||
public int findMin(int[] nums) { | ||
for(int i = 1; i < nums.length; i++) { | ||
if(nums[i - 1] > nums[i]) { | ||
return nums[i]; | ||
} | ||
} | ||
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return nums[0]; | ||
} | ||
} |
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. LinkedList 의 특징을 잘 살려서 풀이 해 주신것 같습니다! |
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Original file line number | Diff line number | Diff line change |
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/** | ||
* Definition for singly-linked list. | ||
* class ListNode { | ||
* int val; | ||
* ListNode next; | ||
* ListNode(int x) { | ||
* val = x; | ||
* next = null; | ||
* } | ||
* } | ||
*/ | ||
//시간복잡도: O(n) | ||
//공간복잡도: O(1) | ||
public class Solution { | ||
public boolean hasCycle(ListNode head) { | ||
ListNode slow = head; | ||
ListNode fast = head; | ||
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while(fast != null && fast.next != null) { | ||
slow = slow.next; | ||
fast = fast.next.next; | ||
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if(slow == fast) return true; | ||
} | ||
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return false; | ||
} | ||
} |
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해당 문제의 경우 시간 복잡도를 O(n) 으로 하셔도 괜찮지만, 조건 상 O(log n) 으로 되어있기 때문에 다시 한번 풀이에 도전해 보셔도 괜찮을것 같아요 :)