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宫水三叶的刷题日记

发布于 2024-06-27 13:24:58

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发布于 2024-06-27 13:24:58

文章被收录于专栏：宫水三叶的刷题日记宫水三叶的刷题日记

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来一道简简单单周三题。

题目描述

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

[x_i, y_i]

，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:

a.length == b.length == c.length == d.length == 2

-10^4 <= x_i, y_i <= 10^4

计算几何

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
}