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来自「浙江大学计算机学院数据结构课程的教学课件」· HTM 代码 · 共 55 行
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<pre align="left"><b><font face="Arial" size="4">
<font color="#FF0033" size="5">OPERATIONS</font>
(1) <font color="#0033CC">transposing a matrix</font>
<font color="#0000FF">algorithm:</font>
<font face="Arial, Helvetica, sans-serif" color="#FF0000">※ </font>for each row i
take element <font face="Arial, Helvetica, sans-serif" color="#FF0000"><i>< i, j , value></i></font> and store it
as element<i> <font face="Arial, Helvetica, sans-serif" color="#FF0000">< j, i , value ></font></i> of the transpose
<font face="Arial, Helvetica, sans-serif" color="#FF0000">※ </font>for all elements in column j
place element<i><font face="Arial, Helvetica, sans-serif" color="#FF0000"> < i, j , value></font></i> and store it
as element<font face="Arial, Helvetica, sans-serif" color="#FF0000"> <i>< j, i , value ></i></font> of the transpose
void <font color="#FF0000" face="Arial, Helvetica, sans-serif">transpose</font> ( term a[ ], term b [ ])
<font color="#CC0099" size="3">/* b is set to the transpose of a */</font>
(
int n, i, j, currentb;
n = a[ 0 ].value; <font color="#CC0099" face="Arial, Helvetica, sans-serif"><i> /* total number of elements */</i></font>
b[0].row = a[0].col ; <i><font face="Arial, Helvetica, sans-serif" size="3" color="#CC0099"> /* rows in b = column in a*/</font></i>
b[0].col = a[0].row ; <font color="#CC0099" size="3"> /* column in b = rows in a */</font>
b[0].value =n;
if (n > 0) { <font color="#CC0099" size="3"> /* non zero matrix*/</font>
currentb = 1;
for ( i =0; i < a[0].col; i++) <i><font face="Arial, Helvetica, sans-serif"> <font color="#CC0099">/* transpose by the column in a*/</font></font></i>
for ( j = 1; j <=n; j++) <font color="#CC0099" size="3"> <i><font face="Arial, Helvetica, sans-serif" size="4"> /* find elements from the current*/</font></i></font>
if (a[j].col == i) {
<i><font face="Arial, Helvetica, sans-serif"> <font color="#CC0099">/* element is in current column, add it to b */</font></font></i>
b[currentb].row =a[j].col;
b[currentb].col =a[j].row;
b[currentb].value =a[j].value;
currentb++;
}
}
}
</font></b></pre>
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