{ "data": { "question": { "questionId": "699", "questionFrontendId": "699", "categoryTitle": "Algorithms", "boundTopicId": 1824, "title": "Falling Squares", "titleSlug": "falling-squares", "content": "
There are several squares being dropped onto the X-axis of a 2D plane.
\n\nYou are given a 2D integer array positions
where positions[i] = [lefti, sideLengthi]
represents the ith
square with a side length of sideLengthi
that is dropped with its left edge aligned with X-coordinate lefti
.
Each square is dropped one at a time from a height above any landed squares. It then falls downward (negative Y direction) until it either lands on the top side of another square or on the X-axis. A square brushing the left/right side of another square does not count as landing on it. Once it lands, it freezes in place and cannot be moved.
\n\nAfter each square is dropped, you must record the height of the current tallest stack of squares.
\n\nReturn an integer array ans
where ans[i]
represents the height described above after dropping the ith
square.
\n
Example 1:
\n\nInput: positions = [[1,2],[2,3],[6,1]]\nOutput: [2,5,5]\nExplanation:\nAfter the first drop, the tallest stack is square 1 with a height of 2.\nAfter the second drop, the tallest stack is squares 1 and 2 with a height of 5.\nAfter the third drop, the tallest stack is still squares 1 and 2 with a height of 5.\nThus, we return an answer of [2, 5, 5].\n\n\n
Example 2:
\n\n\nInput: positions = [[100,100],[200,100]]\nOutput: [100,100]\nExplanation:\nAfter the first drop, the tallest stack is square 1 with a height of 100.\nAfter the second drop, the tallest stack is either square 1 or square 2, both with heights of 100.\nThus, we return an answer of [100, 100].\nNote that square 2 only brushes the right side of square 1, which does not count as landing on it.\n\n\n
\n
Constraints:
\n\n1 <= positions.length <= 1000
1 <= lefti <= 108
1 <= sideLengthi <= 106
在二维平面上的 x 轴上,放置着一些方块。
\n\n给你一个二维整数数组 positions
,其中 positions[i] = [lefti, sideLengthi]
表示:第 i
个方块边长为 sideLengthi
,其左侧边与 x 轴上坐标点 lefti
对齐。
每个方块都从一个比目前所有的落地方块更高的高度掉落而下。方块沿 y 轴负方向下落,直到着陆到 另一个正方形的顶边 或者是 x 轴上 。一个方块仅仅是擦过另一个方块的左侧边或右侧边不算着陆。一旦着陆,它就会固定在原地,无法移动。
\n\n在每个方块掉落后,你必须记录目前所有已经落稳的 方块堆叠的最高高度 。
\n\n返回一个整数数组 ans
,其中 ans[i]
表示在第 i
块方块掉落后堆叠的最高高度。
\n\n
示例 1:
\n\n输入:positions = [[1,2],[2,3],[6,1]]\n输出:[2,5,5]\n解释:\n第 1 个方块掉落后,最高的堆叠由方块 1 组成,堆叠的最高高度为 2 。\n第 2 个方块掉落后,最高的堆叠由方块 1 和 2 组成,堆叠的最高高度为 5 。\n第 3 个方块掉落后,最高的堆叠仍然由方块 1 和 2 组成,堆叠的最高高度为 5 。\n因此,返回 [2, 5, 5] 作为答案。\n\n\n
示例 2:
\n\n\n输入:positions = [[100,100],[200,100]]\n输出:[100,100]\n解释:\n第 1 个方块掉落后,最高的堆叠由方块 1 组成,堆叠的最高高度为 100 。\n第 2 个方块掉落后,最高的堆叠可以由方块 1 组成也可以由方块 2 组成,堆叠的最高高度为 100 。\n因此,返回 [100, 100] 作为答案。\n注意,方块 2 擦过方块 1 的右侧边,但不会算作在方块 1 上着陆。\n\n\n
\n\n
提示:
\n\n1 <= positions.length <= 1000
1 <= lefti <= 108
1 <= sideLengthi <= 106
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