[LeetCode] 223. Rectangle Area

Find the total area covered by two rectilinear rectangles in a 2D plane.

Each rectangle is defined by its bottom left corner and top right corner as shown in the figure.

Example:

Input: A = -3, B = 0, C = 3, D = 4, E = 0, F = -1, G = 9, H = 2
Output: 45

Note:

Assume that the total area is never beyond the maximum possible value of int.

矩形面积。题意是给两个矩形的左下角的坐标和右上角的坐标，请你返回由这两个矩形组成的多边形的面积。

思路是先计算两个矩形各自的面积，再计算两者重叠的部分，最后用两个矩形各自的面积 - 两者重叠的部分即可。各自计算两个长方形的面积这个很好处理，但是如何计算重叠部分的面积呢？这个题不要想复杂了，可以就参照题目给的例子来计算重叠部分的面积。对于这个重叠的部分，左下角的横坐标是A和E的较大值，右上角的横坐标是C和G的较小值，所以重叠部分的长是Math.min(C, G) - Math.max(A, E)。同理，重叠部分的高 = Math.min(D, H) - Math.max(B, F)。

时间O(1)

空间O(1)

Java实现

class Solution {
    public int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
        int areaA = (C - A) * (D - B);
        int areaB = (G - E) * (H - F);
        int left = Math.max(A, E);
        int right = Math.min(C, G);
        int top = Math.min(D, H);
        int bottom = Math.max(B, F);

        int overlap = 0;
        if (right > left && top > bottom) {
            overlap = (right - left) * (top - bottom);
        }
        return areaA + areaB - overlap;
    }
}

JavaScript实现

/**
 * @param {number} A
 * @param {number} B
 * @param {number} C
 * @param {number} D
 * @param {number} E
 * @param {number} F
 * @param {number} G
 * @param {number} H
 * @return {number}
 */
var computeArea = function (A, B, C, D, E, F, G, H) {
    let areaA = (C - A) * (D - B);
    let areaB = (G - E) * (H - F);
    let left = Math.max(A, E);
    let right = Math.min(C, G);
    let top = Math.min(D, H);
    let bottom = Math.max(B, F);

    let overlap = 0;
    if (right > left && top > bottom) {
        overlap = (right - left) * (top - bottom);
    }
    return areaA + areaB - overlap;
};

LeetCode 题目总结