LWC 54：699. Falling Squares
LWC 54：699. Falling Squares
Problem:
On an infinite number line (x-axis), we drop given squares in the order they are given. The i-th square dropped (positions[i] = (left, side_length)) is a square with the left-most point being positions[i][0] and sidelength positionsi. The square is dropped with the bottom edge parallel to the number line, and from a higher height than all currently landed squares. We wait for each square to stick before dropping the next. The squares are infinitely sticky on their bottom edge, and will remain fixed to any positive length surface they touch (either the number line or another square). Squares dropped adjacent to each other will not stick together prematurely. Return a list ans of heights. Each height ans[i] represents the current highest height of any square we have dropped, after dropping squares represented by positions[0], positions1, …, positions[i].
Example 1:
Input: [[1, 2], [2, 3], [6, 1]] Output: [2, 5, 5] Explanation: After the first drop of positions[0] = [1, 2]: _aa _aa The maximum height of any square is 2. After the second drop of positions1 = [2, 3]: __aaa __aaa __aaa aa_ aa_ The maximum height of any square is 5. The larger square stays on top of the smaller square despite where its center of gravity is, because squares are infinitely sticky on their bottom edge. After the third drop of positions1 = [6, 1]: __aaa __aaa __aaa _aa _aa___a The maximum height of any square is still 5. Thus, we return an answer of [2, 5, 5].
Example 2:
Input: [[100, 100], [200, 100]] Output: [100, 100] Explanation: Adjacent squares don’t get stuck prematurely - only their bottom edge can stick to surfaces.
Note:
1 <= positions.length <= 1000.
1 <= positions0 <= 10^8.
1 <= positions1 <= 10^6.
思路： 有点几何题的味道，实际上正方形往上叠总共就5种情况，如下：
前四种情况，可以合并成一种情况处理，只需要搜索蓝色两条边界内是否存在对应的边，如果有，则说明需要叠加。第五种情况需要特殊处理，直接扫描所有正方形，包含蓝色矩形块则更新高度。
代码如下：
class Edge{
int x;
int y;
int h;
Edge(int x, int y, int h){
this.x = x;
this.y = y;
this.h = h;
}
}
public List<Integer> fallingSquares(int[][] positions) {
int n = positions.length;
List<Integer> ans = new ArrayList<>();
TreeMap<Double, Integer> map = new TreeMap<>();
List<Edge> heights = new ArrayList<Edge>();
int max = 0;
for (int i = 0; i < n; ++i) {
int x = positions[i][0];
int h = positions[i][1];
int y = h + x - 1;
int h_max = 0;
for (Double e : map.subMap(x - 0.1, y + 0.1).keySet()) {
h_max = Math.max(h_max, map.get(e));
}
for (Edge edge : heights) {
if (edge.x <= x && edge.y >= y) {
h_max = Math.max(h_max, edge.h);
}
}
h_max += h;
map.put(x * 1.0, h_max);
map.put(y * 1.0, h_max);
heights.add(new Edge(x, y, h_max));
max = Math.max(max, h_max);
ans.add(max);
}
return ans;
} 其实吧，上述五种情况也能合并在一块，比较取巧，先把对应的所有边都枚举出来，之后不断在其基础上累加高度即可。造房子的感觉！
代码如下：
public List<Integer> fallingSquares(int[][] positions) {
int n = positions.length;
List<Integer> ans = new ArrayList<>();
TreeMap<Double, Integer> map = new TreeMap<>();
for (int[] pos : positions) {
int x = pos[0];
int h = pos[1];
int y = x + h - 1;
map.put(x * 1.0, 0);
map.put(y * 1.0, 0);
}
int max = 0;
for (int i = 0; i < n; ++i) {
int x = positions[i][0];
int h = positions[i][1];
int y = h + x - 1;
int h_max = 0;
for (Double e : map.subMap(x - 0.1, y + 0.1).keySet()) {
h_max = Math.max(h_max, map.get(e));
}
h_max += h;
for (Double e : map.subMap(x - 0.1, y + 0.1).keySet()) {
map.put(e, h_max);
}
max = Math.max(max, h_max);
ans.add(max);
}
return ans;
} 腾讯云开发者
