如何在JAVA中按照两个条件进行排序?

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英文:

How to sort by two criteria in JAVA?

问题

我有一个包含以下数据的ArrayList:coordX,coordY,length,position(position可以是3种类型之一:vertical,horizontal或1x1)。使用插入方法,我按长度降序对其进行排序。如果长度相等,如何给予具有水平位置的长度值优先权。对于给定的输入:

(1;0),长度=2,位置-垂直
(1;6),长度=4,位置-水平
(3;4),长度=3,位置-水平
(3;6),长度=1,位置-1x1
(4;0),长度=1,位置-1x1
(5;3),长度=2,位置-水平
(6;7),长度=2,位置-垂直
(7;5),长度=2,位置-水平

输出应为:

(1;6),长度=4,位置-水平
(3;4),长度=3,位置-水平
(5;3),长度=2,位置-水平
(7;5),长度=2,位置-水平
(1;0),长度=2,位置-垂直
(6;7),长度=2,位置-垂直
(3;6),长度=1,位置-1x1
(4;0),长度=1,位置-1x1

目前我的代码如下:

public static void insertionSort(ArrayList<sort> srt) {
    int i, j;
    for (i = 1; i < srt.size(); i++) {
        sort tmp = srt.get(i);
        j = i;
        while ((j > 0) && (srt.get(j - 1).length < tmp.length)) {
            srt.set(j, srt.get(j - 1));
            j--;
        }
        srt.set(j, tmp);
    }
    for (sort e : srt) {
        System.out.println("(" + e.coordX + ";" + e.coordY + ")" + "  ,  length= " + e.length + "  ,  position - " + e.pozitie);
    }
}

此代码部分仅负责按长度排序。

英文:

I got an ArrayList which contains next data: coordX , coordY , length , position ( position can be of 3 types : vertical , horizontal or 1x1 ). Using insertion method , I sort this descending by length . How can i give priority in case if length is equal for that value of length who has the horizontal position.For given input :

(1;0)  ,  length = 2  ,  position - vertical
(1;6)  ,  length = 4  ,  position - horizontal
(3;4)  ,  length = 3  ,  position - horizontal
(3;6)  ,  length = 1  ,  position - 1x1
(4;0)  ,  length = 1  ,  position - 1x1
(5;3)  ,  length = 2  ,  position - horizontal
(6;7)  ,  length = 2  ,  position - vertical
(7;5)  ,  length = 2  ,  position - horizontal

the output should be :

(1;6)  ,  length = 4  ,  position - horizontal
(3;4)  ,  length = 3  ,  position - horizontal
(5;3)  ,  length = 2  ,  position - horizontal
(7;5)  ,  length = 2  ,  position - horizontal
(1;0)  ,  length = 2  ,  position - vertical
(6;7)  ,  length = 2  ,  position - vertical
(3;6)  ,  length = 1  ,  position - 1x1
(4;0)  ,  length = 1  ,  position - 1x1

This is what i have at the moment :

public static void insertionSort(ArrayList&lt;sort&gt; srt) {
		int i,j;
	    for (i = 1; i &lt; srt.size(); i++) {
	        sort tmp = srt.get(i);
	        j = i;
	        while ((j &gt; 0) &amp;&amp; (srt.get(j - 1).length&lt; tmp.length)) {
	            srt.set(j, srt.get(j - 1));
	            j--;
	        }
	        srt.set(j, tmp);
	    }
	    for(sort e : srt) {
	    	System.out.println(&quot;(&quot;+e.coordX+&quot;;&quot;+e.coordY+&quot;)&quot;+&quot;  ,  length= &quot;+e.length+&quot;  ,  position - &quot;+e.pozitie);
	    }

}

This part of code is responsabile only for sorting by the length.

答案1

得分: 1

请尝试以下内容:

您首先需要检查长度,然后如果它们相等,请检查位置。

			while ((j > 0) && (srt.get(j - 1).length < tmp.length
					|| (srt.get(j - 1).length == tmp.length
							&& srt.get(j - 1).position
									.compareTo(tmp.position) > 0))) {
				srt.set(j, srt.get(j - 1));
				j--;
			}
英文:

Try the following:

You need to check first on the length and then if they are equal, check on the position.

			while ((j &gt; 0) &amp;&amp; (srt.get(j - 1).length &lt; tmp.length
					|| (srt.get(j - 1).length == tmp.length
							&amp;&amp; srt.get(j - 1).position
									.compareTo(tmp.position) &gt; 0))) {
				srt.set(j, srt.get(j - 1));
				j--;
			}

</details>



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  • 本文由 发表于 2020年4月8日 05:46:02
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