回归主线。
来一道简简单单周三题。
平台:LeetCode
题号:593
给定 2D
空间中四个点的坐标 a
, b
, c
和 d
,如果这四个点构成一个正方形,则返回 true
。
点的坐标表示为
,输入不是按任何顺序给出的。
一个有效的正方形有四条等边和四个等角(90
度角)。
示例 1:
输入: a = [0,0], b = [1,1], c = [1,0], d = [0,1]
输出: True
示例 2:
输入:a = [0,0], b = [1,1], c = [1,0], d = [0,12]
输出:false
示例 3:
输入:a = [1,0], b = [-1,0], c = [0,1], d = [0,-1]
输出:true
提示:
根据题意进行模拟即可。
从给定的 4 个顶点中选 3 个顶点,检查其能否形成「直角三角形」,同时保存下来首个直角三角形的直角边边长,供后续其余直角三角形进行对比(注意不能共点,即直角边长不能为 0)。
Java 代码:
class Solution {
long len = -1;
public boolean validSquare(int[] a, int[] b, int[] c, int[] d) {
return calc(a, b, c) && calc(a, b, d) && calc(a, c, d) && calc(b, c, d);
}
boolean calc(int[] a, int[] b, int[] c) {
long l1 = (a[0] - b[0]) * (a[0] - b[0]) + (a[1] - b[1]) * (a[1] - b[1]);
long l2 = (a[0] - c[0]) * (a[0] - c[0]) + (a[1] - c[1]) * (a[1] - c[1]);
long l3 = (b[0] - c[0]) * (b[0] - c[0]) + (b[1] - c[1]) * (b[1] - c[1]);
boolean ok = (l1 == l2 && l1 + l2 == l3) || (l1 == l3 && l1 + l3 == l2) || (l2 == l3 && l2 + l3 == l1);
if (!ok) return false;
if (len == -1) len = Math.min(l1, l2);
else if (len == 0 || len != Math.min(l1, l2)) return false;
return true;
}
}
C++ 代码:
class Solution {
public:
long len = -1;
bool validSquare(vector<int>& a, vector<int>& b, vector<int>& c, vector<int>& d) {
return calc(a, b, c) && calc(a, b, d) && calc(a, c, d) && calc(b, c, d);
}
bool calc(vector<int>& a, vector<int>& b, vector<int>& c) {
long l1 = (a[0] - b[0]) * (a[0] - b[0]) + (a[1] - b[1]) * (a[1] - b[1]);
long l2 = (a[0] - c[0]) * (a[0] - c[0]) + (a[1] - c[1]) * (a[1] - c[1]);
long l3 = (b[0] - c[0]) * (b[0] - c[0]) + (b[1] - c[1]) * (b[1] - c[1]);
bool ok = (l1 == l2 && l1 + l2 == l3) || (l1 == l3 && l1 + l3 == l2) || (l2 == l3 && l2 + l3 == l1);
if (!ok) return false;
if (len == -1) len = min(l1, l2);
else if (len == 0 || len != min(l1, l2)) return false;
return true;
}
};
Python 代码:
class Solution:
def validSquare(self, a: list, b: list, c: list, d: list) -> bool:
length = -1
def calc(a, b, c):
nonlocal length
l1 = (a[0] - b[0]) ** 2 + (a[1] - b[1]) ** 2
l2 = (a[0] - c[0]) ** 2 + (a[1] - c[1]) ** 2
l3 = (b[0] - c[0]) ** 2 + (b[1] - c[1]) ** 2
ok = (l1 == l2 and l1 + l2 == l3) or (l1 == l3 and l1 + l3 == l2) or (l2 == l3 and l2 + l3 == l1)
if not ok: return False
if length == -1: length = min(l1, l2)
elif length == 0 or length != min(l1, l2): return False
return True
return calc(a, b, c) and calc(a, b, d) and calc(a, c, d) and calc(b, c, d)
TypeScript 代码:
let len = -1
function validSquare(a: number[], b: number[], c: number[], d: number[]): boolean {
len = -1
return calc(a, b, c) && calc(a, b, d) && calc(a, c, d) && calc(b, c, d)
};
function calc(a: number[], b: number[], c: number[]): boolean {
const l1 = (a[0] - b[0]) * (a[0] - b[0]) + (a[1] - b[1]) * (a[1] - b[1])
const l2 = (a[0] - c[0]) * (a[0] - c[0]) + (a[1] - c[1]) * (a[1] - c[1])
const l3 = (b[0] - c[0]) * (b[0] - c[0]) + (b[1] - c[1]) * (b[1] - c[1])
const ok = (l1 == l2 && l1 + l2 == l3) || (l1 == l3 && l1 + l3 == l2) || (l2 == l3 && l2 + l3 == l1)
if (!ok) return false
if (len == -1) len = Math.min(l1, l2)
else if (len == 0 || len != Math.min(l1, l2)) return false
return true
}