|
| 1 | +## 每日一题 - 593. 有效的正方形 |
| 2 | + |
| 3 | +### 信息卡片 |
| 4 | + |
| 5 | +- 时间:2019-06-08 |
| 6 | +- 题目链接:https://leetcode.com/problems/top-k-frequent-elements/description/ |
| 7 | +- tag:`Hash Table` `Heap` |
| 8 | + |
| 9 | +### 题目描述 |
| 10 | + |
| 11 | +``` |
| 12 | +Given a non-empty array of integers, return the k most frequent elements. |
| 13 | + |
| 14 | +Example 1: |
| 15 | + |
| 16 | +Input: nums = [1,1,1,2,2,3], k = 2 |
| 17 | +Output: [1,2] |
| 18 | +
|
| 19 | +Example 2: |
| 20 | +
|
| 21 | +
|
| 22 | +Input: nums = [1], k = 1 |
| 23 | +Output: [1] |
| 24 | + |
| 25 | +Note: |
| 26 | + |
| 27 | +You may assume k is always valid, 1 ≤ k ≤ number of unique elements. |
| 28 | +Your algorithm's time complexity must be better than O(n log n), where n is |
| 29 | +the array's size. |
| 30 | +``` |
| 31 | + |
| 32 | + |
| 33 | + |
| 34 | +### 参考答案 |
| 35 | + |
| 36 | +模仿 [@raof01](https://github.com/raof01) 的思路写的JS代码, |
| 37 | + |
| 38 | +基本思路就是: 证明四个角都是直角, 而证明直角的方式就是边长关系。 |
| 39 | + |
| 40 | +四个点一共有六个连接的线段,其中两个是对角线,另外四个是边。 |
| 41 | + |
| 42 | +对于直角来说,满足“a * a + b * b = c * c”, 由于是正方形,所以a = b, 因此c就等于 |
| 43 | +2 * a * a , 其中a为边长,c就是对角线的长度。 |
| 44 | + |
| 45 | + |
| 46 | +我们分别计算出距离的平方,如果有四个相同,另外两个相同。 且二者的关系可以满足直角,那么他就有四个直角,他就是一个正方形 |
| 47 | + |
| 48 | +```js |
| 49 | +/* |
| 50 | + * @lc app=leetcode id=593 lang=javascript |
| 51 | + * |
| 52 | + * [593] Valid Square |
| 53 | + */ |
| 54 | +function square(p1, p2) { |
| 55 | + const deltaX = p1[0] - p2[0]; |
| 56 | + const deltaY = p1[1] - p2[1]; |
| 57 | + |
| 58 | + return deltaX * deltaX + deltaY * deltaY; |
| 59 | +} |
| 60 | +/** |
| 61 | + * @param {number[]} p1 |
| 62 | + * @param {number[]} p2 |
| 63 | + * @param {number[]} p3 |
| 64 | + * @param {number[]} p4 |
| 65 | + * @return {boolean} |
| 66 | + */ |
| 67 | +var validSquare = function (p1, p2, p3, p4) { |
| 68 | + // 证明四个角都是直角 |
| 69 | + // 证明直角的方式就是边长关系 |
| 70 | + const squares = [ |
| 71 | + square(p1, p2), |
| 72 | + square(p1, p3), |
| 73 | + square(p1, p4), |
| 74 | + square(p2, p3), |
| 75 | + square(p2, p4), |
| 76 | + square(p3, p4) |
| 77 | + ]; |
| 78 | + let cnt1 = 0; |
| 79 | + let cnt2 = 0; |
| 80 | + let sum = 0; |
| 81 | + |
| 82 | + for(let i = 0; i < squares.length; i++) { |
| 83 | + sum += squares[i]; |
| 84 | + } |
| 85 | + |
| 86 | + for(let i = 0; i < squares.length; i++) { |
| 87 | + if (sum === 8 * squares[i]) { |
| 88 | + cnt1++; |
| 89 | + } else if(sum === 4 * squares[i]) { |
| 90 | + cnt2++; |
| 91 | + } |
| 92 | + } |
| 93 | + |
| 94 | + return cnt1 === 4 && cnt2 ===2; |
| 95 | + |
| 96 | +} |
| 97 | +``` |
| 98 | +### 其他优秀解答 |
| 99 | + |
| 100 | +暂无 |
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