95 lines
2.6 KiB
Java
95 lines
2.6 KiB
Java
//给定平面上 n 对 互不相同 的点 points ，其中 points[i] = [xi, yi] 。回旋镖 是由点 (i, j, k) 表示的元组 ，其中
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// i 和 j 之间的距离和 i 和 k 之间的距离相等（需要考虑元组的顺序）。
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//
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// 返回平面上所有回旋镖的数量。
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//
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//
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// 示例 1：
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//
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//
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//输入：points = [[0,0],[1,0],[2,0]]
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//输出：2
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//解释：两个回旋镖为 [[1,0],[0,0],[2,0]] 和 [[1,0],[2,0],[0,0]]
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//
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//
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// 示例 2：
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//
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//
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//输入：points = [[1,1],[2,2],[3,3]]
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//输出：2
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//
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//
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// 示例 3：
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//
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//
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//输入：points = [[1,1]]
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//输出：0
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//
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//
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//
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//
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// 提示：
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//
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//
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// n == points.length
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// 1 <= n <= 500
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// points[i].length == 2
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// -10⁴ <= xi, yi <= 10⁴
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// 所有点都 互不相同
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//
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// Related Topics 数组 哈希表 数学 👍 176 👎 0
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package leetcode.editor.cn;
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import com.code.leet.entiy.TwoArray;
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import java.util.Arrays;
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//447:回旋镖的数量
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class NumberOfBoomerangs {
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public static void main(String[] args) {
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//测试代码
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Solution solution = new NumberOfBoomerangs().new Solution();
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TwoArray twoArray = new TwoArray("[[0,0],[1,0],[2,0]]", true);
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System.out.println(solution.numberOfBoomerangs(twoArray.getArr()));
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}
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//力扣代码
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//leetcode submit region begin(Prohibit modification and deletion)
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class Solution {
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public int numberOfBoomerangs(int[][] points) {
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int size = points.length;
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int[][] lengths = new int[size][size];
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for (int i = 0; i < size; i++) {
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for (int j = i + 1; j < size; j++) {
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int x = points[j][0] - points[i][0];
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int y = points[j][1] - points[i][1];
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int sum = x * x + y * y;
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lengths[i][j] = sum;
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lengths[j][i] = sum;
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}
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}
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int counts = 0;
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for (int[] length : lengths) {
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Arrays.sort(length);
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int count = 0;
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int bef = length[0];
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for (int i = 1; i < size; i++) {
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if (length[i] == bef) {
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count++;
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} else {
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counts += (count + 1) * count;
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count = 0;
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bef = length[i];
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}
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}
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if (count > 0) {
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counts += (count + 1) * count;
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}
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}
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return counts;
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}
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}
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//leetcode submit region end(Prohibit modification and deletion)
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} |