[Leetcode] 406. Queue Reconstruction by Height 直覺解釋
阿新 • • 發佈:2018-12-04
406. Queue Reconstruction by Height
https://leetcode.com/problems/queue-reconstruction-by-height/
首先,我們找到最小的 height (h, k)
,這個資料對的最終位置應該是在 k + 1
的位置上。因為這個值是最小的值,所以其它的值就不小於它。
如果它不在 k + 1
這個位置上,比如在 k + 1 + 1
這個位置上,那麼它就應該是 (h, k + 1)
,因為它前面有 k + 1
個數是大於等於它的。
然後,我們看一下次小的值。因為在最終位置上的數都是小於它的,所以最終結果表上的那些資料對它的 k
k + 1
的位置上就行了。
例子:
輸入: [[7,0],[4,4],[7,1],[5,0],[6,1],[5,2]] 排序之後: [[4, 4], [5, 0], [5, 2], [6, 1], [7, 0], [7, 1]] 1. [None, None, None, None, [4, 4], None] 2. [[5, 0], None, None, None, [4, 4], None] 3. [[5, 0], None, [5, 2], None, [4, 4], None] 4. [[5, 0], None, [5, 2], [6, 1], [4, 4], None] 5. [[5, 0], [7, 0], [5, 2], [6, 1], [4, 4], None] 6. [[5, 0], [7, 0], [5, 2], [6, 1], [4, 4], [7, 1]]
實現
class Solution(object): def reconstructQueue(self, people): """ :type people: List[List[int]] :rtype: List[List[int]] """ people.sort() n = len(people) table = [i for i in range(n)] # store the final positions res = [None] * n while people: top = people.pop(0) stack = [top] # find the same height while people and people[0][0] == stack[0][0]: stack.append(people.pop(0)) # put them in the final position while stack: top = stack.pop() idx = top[1] res[table.pop(idx)] = top # put in the final position return res