trianglearray.htm
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<title>上三角、下三角、对称矩阵</title>
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<h3><a href="http://caterpillar.onlyfun.net/GossipCN/index.html">From
Gossip@caterpillar</a></h3>
<h1><a href="AlgorithmGossip.htm">Algorithm Gossip: 上三角、下三角、对称矩阵</a></h1>
<h2>说明</h2>
上三角矩阵是矩阵在对角线以下的元素均为0，即A<sub>ij</sub> = 0，i > j，例如：<br>
<div style="margin-left: 40px;"><span style="font-weight: bold; font-family: Courier New,Courier,monospace;">1 2 3 4 5 </span><br style="font-weight: bold; font-family: Courier New,Courier,monospace;">
<span style="font-weight: bold; font-family: Courier New,Courier,monospace;">0 6 7 8 9 </span><br style="font-weight: bold; font-family: Courier New,Courier,monospace;">
<span style="font-weight: bold; font-family: Courier New,Courier,monospace;">0 0 10 11 12 </span><br style="font-weight: bold; font-family: Courier New,Courier,monospace;">
<span style="font-weight: bold; font-family: Courier New,Courier,monospace;">0 0 0 13 14 </span><br style="font-weight: bold; font-family: Courier New,Courier,monospace;">
<span style="font-weight: bold; font-family: Courier New,Courier,monospace;">0 0 0 0 15</span><br>
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下三角矩阵是矩阵在对角线以上的元素均为0，即A<sub>ij </sub>= 0，i < j，例如：<br>
<div style="margin-left: 40px;"><span style="font-weight: bold; font-family: Courier New,Courier,monospace;"> 1 0 0 0 0 </span><br style="font-weight: bold; font-family: Courier New,Courier,monospace;">
<span style="font-weight: bold; font-family: Courier New,Courier,monospace;"> 2 6 0 0 0 </span><br style="font-weight: bold; font-family: Courier New,Courier,monospace;">
<span style="font-weight: bold; font-family: Courier New,Courier,monospace;"> 3 7 10 0 0 </span><br style="font-weight: bold; font-family: Courier New,Courier,monospace;">
<span style="font-weight: bold; font-family: Courier New,Courier,monospace;"> 4 8 11 13 0 </span><br style="font-weight: bold; font-family: Courier New,Courier,monospace;">
<span style="font-weight: bold; font-family: Courier New,Courier,monospace;"> 5 9 12 14 15 </span><br>
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对称矩阵是矩阵元素对称于对角线，例如：<br>
<div style="margin-left: 40px;"><span style="font-weight: bold; font-family: Courier New,Courier,monospace;"> 1 2 3 4 5 </span><br style="font-weight: bold; font-family: Courier New,Courier,monospace;">
<span style="font-weight: bold; font-family: Courier New,Courier,monospace;"> 2 6 7 8 9 </span><br style="font-weight: bold; font-family: Courier New,Courier,monospace;">
<span style="font-weight: bold; font-family: Courier New,Courier,monospace;"> 3 7 10 11 12 </span><br style="font-weight: bold; font-family: Courier New,Courier,monospace;">
<span style="font-weight: bold; font-family: Courier New,Courier,monospace;"> 4 8 11 13 14 </span><br style="font-weight: bold; font-family: Courier New,Courier,monospace;">
<span style="font-weight: bold; font-family: Courier New,Courier,monospace;"> 5 9 12 14 15 </span><br>
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上三角或下三角矩阵也有大部份的元素不储存值（为0），我们可以将它们使用一维阵列来储存以节省储存空间，而对称矩阵因为对称于对角线，所以可以视为上三角或下三角矩阵来储存。<br>
<h2>解法</h2>
假设矩阵为nxn，为了计算方便，我们让阵列索引由1开始，上三角矩阵化为一维阵列，若以列为主，其公式为：<br>
<div style="margin-left: 40px;"><span style="font-weight: bold; font-family: Courier New,Courier,monospace;">loc = n*(i-1) - i*(i-1)/2 + j</span><br>
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化为以行为主，其公式为：<br>
<div style="margin-left: 40px;"><span style="font-weight: bold; font-family: Courier New,Courier,monospace;">loc = j*(j-1)/2 + i</span><br>
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下三角矩阵化为一维阵列，若以列为主，其公式为：<br>
<div style="margin-left: 40px;"><span style="font-weight: bold; font-family: Courier New,Courier,monospace;">loc = i*(i-1)/2 + j</span><br>
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若以行为主，其公式为：<br>
<div style="margin-left: 40px;"><span style="font-weight: bold; font-family: Courier New,Courier,monospace;">loc = n*(j-1) - j*(j-1)/2 + i</span><br>
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公式的导证其实是由等差级数公式得到，您可以自行绘图并看看就可以导证出来，对于C/C++或Java等索引由0开始的语言来说，只要将i与j各加1，求得loc之后减1即可套用以上的公式。 <br>
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<h2> 实作</h2>
<ul>
<li> C
</li>
</ul>
<pre>#include <stdio.h> <br>#include <stdlib.h> <br>#define N 5 <br><br>int main(void) { <br> int arr1[N][N] = { <br> {1, 2, 3, 4, 5}, <br> {0, 6, 7, 8, 9}, <br> {0, 0, 10, 11, 12}, <br> {0, 0, 0, 13, 14}, <br> {0, 0, 0, 0, 15}}; <br><br> int arr2[N*(1+N)/2] = {0}; <br><br> int i, j, loc = 0; <br><br> printf("原二维资料：\n"); <br> for(i = 0; i < N; i++) { <br> for(j = 0; j < N; j++) { <br> printf("%4d", arr1[i][j]); <br> } <br> printf("\n"); <br> } <br><br> printf("\n以列为主："); <br> for(i = 0; i < N; i++) { <br> for(j = 0; j < N; j++) { <br> if(arr1[i][j] != 0) <br> arr2[loc++] = arr1[i][j]; <br> } <br> } <br> for(i = 0; i < N*(1+N)/2; i++) <br> printf("%d ", arr2[i]); <br><br> printf("\n输入索引(i, j)："); <br> scanf("%d, %d", &i, &j); <br> loc = N*i - i*(i+1)/2 + j; <br> printf("(%d, %d) = %d", i, j, arr2[loc]); <br><br> printf("\n"); <br><br> return 0; <br>} <br></pre>
<br>
<ul>
<li> Java
</li>
</ul>
<pre>public class TriangleArray {<br> private int[] arr;<br> private int length;<br> <br> public TriangleArray(int[][] array) {<br> length = array.length;<br> arr = new int[length*(1+length)/2];<br> <br> int loc = 0;<br> for(int i = 0; i < length; i++) { <br> for(int j = 0; j < length; j++) { <br> if(array[i][j] != 0) <br> arr[loc++] = array[i][j]; <br> } <br> }<br> }<br> <br> public int getValue(int i, int j) {<br> int loc = length*i - i*(i+1)/2 + j;<br> return arr[loc];<br> }<br><br> public static void main(String[] args) {<br> int[][] array = { <br> {1, 2, 3, 4, 5}, <br> {0, 6, 7, 8, 9}, <br> {0, 0, 10, 11, 12}, <br> {0, 0, 0, 13, 14}, <br> {0, 0, 0, 0, 15}};<br> <br> TriangleArray triangleArray =<br> new TriangleArray(array);<br> <br> System.out.print(triangleArray.getValue(2, 2));<br> }<br>}</pre>
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