牛客网:为什么不能将实数作为 HashMap 的 key?
程序员大白
共 565字,需浏览 2分钟
·
2022-01-22 16:20
点击上方“程序员大白”,选择“星标”公众号
重磅干货,第一时间送达
1.起因
Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.
import java.util.HashMap;
import java.util.Map;
//class Point {
// int x;
// int y;
// Point(int a, int b) { x = a; y = b; }
//}
public class Solution {
public int maxPoints(Point[] points) {
if (points.length <= 2) {
return points.length;
}
int max = 2;
for (int i = 0; i < points.length - 1; i++) {
Map
map = new HashMap<>(16); // 记录垂直点数; 当前和Points[i]在一条线上的最大点数; 和Points[i]垂直的点数
int ver = 0, cur, dup = 0;
for (int j = i + 1; j < points.length; j++) {
if (points[j].x == points[i].x) {
if (points[j].y != points[i].y) {
ver++;
} else {
dup++;
}
} else {
float d = (float)((points[j].y - points[i].y) / (double) (points[j].x - points[i].x));
map.put(d, map.get(d) == null ? 1 : map.get(d) + 1);
}
}
cur = ver;
for (int v : map.values()) {
cur = Math.max(v, cur);
}
max = Math.max(max, cur + dup + 1);
}
return max;
}
}
public static void main(String[] args) {
int[][] vals = {{2,3},{3,3},{-5,3}};
Point[] points = new Point[3];
for (int i=0; i
points[i] = new Point(vals[i][0], vals[i][1]);
}
Solution solution = new Solution();
System.out.println(solution.maxPoints(points));
}
System.out.println(0.0 == -0.0);
map.put(d, map.get(d) == null ? 1 : map.get(d) + 1);
public V get(Object key) {
Node
e; return (e = getNode(hash(key), key)) == null ? null : e.value;
}
static final int hash(Object key) {
int h;
return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
}
public native int hashCode();
public static void main(String[] args) {
System.out.println(0.0 == -0.0);
System.out.println(new Float(0.0).hashCode() ==
new Float(-0.0).hashCode());
}
在程序执行期间,只要equals方法的比较操作用到的信息没有被修改,那么对这同一个对象调用多次,hashCode方法必须始终如一地返回同一个整数。
如果两个对象根据equals方法比较是相等的,那么调用两个对象的hashCode方法必须返回相同的整数结果。
如果两个对象根据equals方法比较是不等的,则hashCode方法不一定得返回不同的整数。——《effective java》
System.out.println(new Float(0.0).equals(0.0f));
System.out.println(new Float(0.0).equals((float) -0.0));
public boolean equals(Object obj) {
return (obj instanceof Float)
&& (floatToIntBits(((Float)obj).value) ==
floatToIntBits(value));
}
/**
* Returns a representation of the specified floating-point value
* according to the IEEE 754 floating-point "single format" bit
* layout.
*
*
Bit 31 (the bit that is selected by the mask
* {@code 0x80000000}) represents the sign of the floating-point
* number.
* Bits 30-23 (the bits that are selected by the mask
* {@code 0x7f800000}) represent the exponent.
* Bits 22-0 (the bits that are selected by the mask
* {@code 0x007fffff}) represent the significand (sometimes called
* the mantissa) of the floating-point number.
*
*
If the argument is positive infinity, the result is
* {@code 0x7f800000}.
*
*
If the argument is negative infinity, the result is
* {@code 0xff800000}.
*
*
If the argument is NaN, the result is {@code 0x7fc00000}.
*
*
In all cases, the result is an integer that, when given to the
* {@link #intBitsToFloat(int)} method, will produce a floating-point
* value the same as the argument to {@code floatToIntBits}
* (except all NaN values are collapsed to a single
* "canonical" NaN value).
*
* @param value a floating-point number.
* @return the bits that represent the floating-point number.
*/
public static int floatToIntBits(float value) {
int result = floatToRawIntBits(value);
// Check for NaN based on values of bit fields, maximum
// exponent and nonzero significand.
if (((result & FloatConsts.EXP_BIT_MASK) ==
FloatConsts.EXP_BIT_MASK) &&
(result & FloatConsts.SIGNIF_BIT_MASK) != 0)
result = 0x7fc00000;
return result;
}
当浮点运算产生一个非常接近0的负浮点数时,会产生“-0.0”,而这个浮点数不能正常表示
System.out.println(Float.floatToIntBits((float) 0.0));
System.out.println(Float.floatToIntBits((float) -0.0));
System.out.println(Float.floatToRawIntBits(0.0f));
System.out.println(Float.floatToRawIntBits((float)-0.0));
0
-2147483648
0
-2147483648
推荐阅读
关于程序员大白
程序员大白是一群哈工大,东北大学,西湖大学和上海交通大学的硕士博士运营维护的号,大家乐于分享高质量文章,喜欢总结知识,欢迎关注[程序员大白],大家一起学习进步!
评论