c8913be0da89 ·

貌似现在leetcode多了rectangle是个点的check？ 还有一个rectangle inside另外一个rectangle
感觉个人做 写了很多extra if statement来check 不知道有没有更好的解法

public int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
// A to G's right or E to C's right to avoid X
// Similarly B(bottom!!!) >= H or F>= D
if ( (A >= G ) ||(E >= C) ||( B>=H ) ||(F >= D) ){
return (C-A)*(D-B) + (G-E)*(H-F);
}else{
//prevents points
if ((A==C)||(B==D)||(E==G)||(F==H)){
return (A-C)*(B-D) + (E-G)*(F-H) ;
}
// prevents inner within a larger rectangle

int maxLx = A - E > 0 ? A: E;
int maxLy = Math.max(B, F);
int minLx = Math.min(C,G);
int minLy = Math.min(D, H);

// check if inner rectangle
if ( ((E-A) >=0 && (C-G)>=0) && ((F-B)>=0) && ((D-H) >=0)||
( ((A-E)>=0)&& ((G-C)>=0 )&& ((B-F)>=0)&&((H-D)>=0) )
){

return (C-A)*(D-B) - (G-E)*(H-F) >=0 ? (C-A)*(D-B): (G-E)*(H-F);
}
int delta =(maxLx - minLx)*(maxLy - minLy);

return (C-A)*(D-B) + (G-E)*(H-F) - delta;
}

}

c8913be0da89 评论自Overlap Rectangle

c8913be0da89

貌似现在leetcode多了rectangle是个点的check？ 还有一个rectangle inside另外一个rectangle
感觉个人做 写了很多extra if statement来check 不知道有没有更好的解法

public int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
// A to G's right or E to C's right to avoid X
// Similarly B(bottom!!!) >= H or F>= D
if ( (A >= G ) ||(E >= C) ||( B>=H ) ||(F >= D) ){
return (C-A)*(D-B) + (G-E)*(H-F);
}else{
//prevents points
if ((A==C)||(B==D)||(E==G)||(F==H)){
return (A-C)*(B-D) + (E-G)*(F-H) ;
}
// prevents inner within a larger rectangle

int maxLx = A - E > 0 ? A: E;
int maxLy = Math.max(B, F);
int minLx = Math.min(C,G);
int minLy = Math.min(D, H);

// check if inner rectangle
if ( ((E-A) >=0 && (C-G)>=0) && ((F-B)>=0) && ((D-H) >=0)||
( ((A-E)>=0)&& ((G-C)>=0 )&& ((B-F)>=0)&&((H-D)>=0) )
){

return (C-A)*(D-B) - (G-E)*(H-F) >=0 ? (C-A)*(D-B): (G-E)*(H-F);
}
int delta =(maxLx - minLx)*(maxLy - minLy);

return (C-A)*(D-B) + (G-E)*(H-F) - delta;
}

}

Overlap Rectangle

这道题与Leetcode 223题RectangleArea十分相似。 题目 给定两个长方形左下角和右上角的坐标，判断是否有重叠，返回true或者false。 Java 代码...

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