題目

https://leetcode.com/problems/rectangle-area-ii/

題解

沒有看懂官方答案，評論區有一種解法寫的挺通俗的：

Clean Recursive Solution [Java]

The idea here is to maintain a list of non-overlapping rectangles to calculate final area.
If a new rectangle does not overlap with any of the existing rectanlges, add it to the list.
If there is an overlap, split the non-overlapping regions of the rectangle into smaller rectangles and compare with the rest of the list.

For example, when a new rectangle (green) is compared with the current rectangle (blue), the non-overlapping regions can be split into two smaller rectangles. Rectangle 1 will be covered by the first overlapping case in addRectangle() and rectangle 2 will be covered by the third case. Rectangle 3 overlaps with the current rectangle and need not be considered.

看了之后自己又實現了一遍：

class Solution {// 0左 1下 2右 3上public static final int L = 0;public static final int D = 1;public static final int R = 2;public static final int U = 3;public int rectangleArea(int[][] rectangles) {ArrayList<int[]> list = new ArrayList<>();for (int[] cur : rectangles) {addRec(cur, list, 0);}long sum = 0;for (int[] r : list) {sum += ((long) r[U] - r[D]) * (r[R] - r[L]) % 1000000007;}return (int) (sum % 1000000007);}public void addRec(int[] cur, ArrayList<int[]> list, int index) {// 全部沒有重疊if (index >= list.size()) {list.add(cur);return;}int[] r = list.get(index);// 當前沒有重疊: cur右邊界小于r左邊界|cur左邊界大于r右邊界|cur下邊界大于r上邊界|cur上邊界小于r下邊界if (cur[R] <= r[L] || cur[L] >= r[R] || cur[D] >= r[U] || cur[U] <= r[D]) {addRec(cur, list, index + 1);return;}// cur的左邊界更左if (cur[L] < r[L]) addRec(new int[]{cur[L], cur[D], r[L], cur[U]}, list, index + 1);// cur的右邊界更右if (cur[R] > r[R]) addRec(new int[]{r[R], cur[D], cur[R], cur[U]}, list, index + 1);// cur的下邊界更下if (cur[D] < r[D]) addRec(new int[]{Math.max(cur[L], r[L]), cur[D], Math.min(cur[R], r[R]), r[D]}, list, index + 1);// cur的上邊界更上if (cur[U] > r[U]) addRec(new int[]{Math.max(cur[L], r[L]), r[U], Math.min(cur[R], r[R]), cur[U]}, list, index + 1);} }

總結

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