[leetcode] 223. Rectangle Area

最新推荐文章于 2025-08-09 17:01:50 发布

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最新推荐文章于 2025-08-09 17:01:50 发布

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分类专栏： leetcode 文章标签： leetcode

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本文介绍了一种计算两个矩形在二维平面上覆盖总面积的方法。通过计算两个矩形的总面积并减去重叠部分得到最终结果。文章提供了两种实现方案，一种简洁高效，另一种详细列举了所有可能的情况。

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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.

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

这道题给出两个矩形，计算他们覆盖的总面积，题目难度为easy。

很容易想到，计算两个矩形的总面积然后减去重叠的部分就得到了总面积，关键在于重叠部分面积的计算。具体代码：

class Solution {
public:
    int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
        int area = 0;
        area += (C - A) * (D - B);
        area += (G - E) * (H - F);
        if(C<=E || A>=G || D<=F || B>=H) return area;
        else return area - (min(C, G) - max(A, E)) * (min(D, H) - max(B, F));
    }
};

查看别人代码时，看到一条讨论标题是“If you want to laugh, look at my solution”，不得不承认人才确实多，忍不住笑了，给他点一个大大的赞，代码：

int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
    int area1 = (D - B)*(C - A);
    int area2 = (H - F)*(G - E);
    int area3;
    if (area1 == 0) {
        return area2;
    }
    if (area2 == 0) {
        return area1;
    }
    if ((A == D) && (B == F) && (C == G) && (D == H)) {
        return area1;
    }
    if ((E >= C) | (G <= A) | (H <= B) | (D <= F)) {    //not overlapping
        return (area1 + area2);
    }
    if (((G - E) <= (C - A)) && ((H - F) <= (D - B)) && (E >= A) && (F >= B) && (G <= C) && (D >= H)) {                        //rect2 is inside rect1
        return area1;
    }
    if (((C - A) <= (G - E)) && ((D - B) <= (H - F)) && (E <= A) && (B >= F) && (G >= C) && (H >= D)) {                        //rect1 is inside rect2
        return area2;
    }
    if ((F >= B) && (E >= A) && (G >= C) && (H >= D)) {                       //overlapping upper right corner
        area3 = (C - E)*(D - F);
    }
    else if ((F >= B) && (E <= A) && (G <= C) && (H >= D)) {                       //overlapping upper left corner
        area3 = (G - A)*(D - F);
    }
    else if ((F <= B) && (E <= A) && (G <= C) && (H <= D)) {                       //overlapping bottom left corner
        area3 = (G - A)*(H - B);
    }
    else if ((F <= B) && (E >= A) && (G >= C) && (H <= D)) {                        //overlapping bottom right corner
        area3 = (H - B)*(C - E);
    }
    else if (((C - A) <= (G - E)) && (H <= D) && (G >= C) && (E <= A) && (F <= B)) {               //overlapping bottom side
        area3 = (C - A)*(H - B);
    }
    else if (((C - A) <= (G - E)) && (H >= D) && (G >= C) && (E <= A) && (F >= B)) {               //overlapping top side
        area3 = (C - A)*(D - F);
    }
    else if (((D - B) <= (H - F)) && (E <= A) && (F <= B) && (H >= D) && (G <= C)) {               //overlapping left side
        area3 = (G - A)*(D - B);
    }
    else if (((D - B) <= (H - F)) && (E >= A) && (F <= B) && (H >= D) && (G >= C)) {               //overlapping right side
        area3 = (C - E)*(D - B);
    }
    else if (((C - A) >= (G - E)) && (E >= A) && (F >= B) && (C >= G) && (D <= H)) {      //overlapping part of top side
        area3 = (G - E)*(D - F);
    }
    else if (((C - A) >= (G - E)) && (A <= E) && (B >= F) && (G <= C) && (D >= H)) {       //overlapping part of bottom side
        area3 = (G - E)*(H - B);
    }
    else if (((D - B) >= (H - F)) && (E <= A) && (F >= B) && (G <= C) && (H <= D)) {      //overlapping part of left side
        area3 = (G - A)*(H - F);
    }
    else if (((D - B) >= (H - F)) && (E >= A) && (F >= B) && (G >= C) && (H <= D)) {       //overlapping part of right side
        area3 = (C - E)*(H - F);
    }
    else if (((G - E) <= (C - A)) && (E >= A) && (F <= B) && (G <= C) && (H >= D)) {     //overlapping top and bottom
        area3 = (G - E)*(D - B);
    }
    else if (((H - F) <= (D - B)) && (E <= A) && (F >= B) && (C <= G) && (D >= H)) {     //overlapping left and right
        area3 = (C - A)*(H - F);
    }

    return (area1 + area2 - area3);
}