Java for LeetCode 149 Max Points on a Line

標籤：

Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.

解題思路：

本題主要需要考慮到斜線的情況，可以分別計算出過points[i]直線最多含幾個點，然後算出最大即可，由於計算points[i]的時候，前面的點都計算過了，所以不需要把前面的點考慮進去，所以問題可以轉化為過points[i]的直線最大點的個數，解題思路是用一個HashMap儲存斜率，遍曆points[i]後面每個點，看看和points[i]的斜率是否出現過，這裡有個問題，如何儲存斜率，理想狀況下，斜率應該用一個String表示，如(0,0)和(2，4)可以儲存為"1k2"這樣的類型，這會涉及到求最大公約數的問題，實現起來比較麻煩，實際上直接用double表示即可通過，這是因為((double)1/((double)Integer.MAX_VALUE-(double)Integer.MIN_VALUE))是能輸出2.3283064370807974E-10的。因此，我們直接用double即可。

JAVA實現如下：

    public int maxPoints(Point[] points) {if (points.length <= 2)return points.length;int max = 2;for (int i = 0; i < points.length; i++) {int pointMax = 1, samePointCount = 0;HashMap<Double, Integer> slopeCount = new HashMap<Double, Integer>();Point origin = points[i];for (int j = i + 1; j < points.length; j++) {Point target = points[j];if (origin.x == target.x && origin.y == target.y) {samePointCount++;continue;}double k;if (origin.x == target.x)k = Float.POSITIVE_INFINITY;else if (origin.y == target.y)k = 0;elsek = ((double) (origin.y - target.y))/ (double) (origin.x - target.x);if (slopeCount.containsKey(k))slopeCount.put(k, slopeCount.get(k) + 1);elseslopeCount.put(k, 2);pointMax = Math.max(pointMax, slopeCount.get(k));}max = Math.max(max, pointMax + samePointCount);}return max;      }

 

Java for LeetCode 149 Max Points on a Line