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<b><font face="Arial, Helvetica, sans-serif" size="4" color="#000000">
<font color="#FF0000" size="5">7.7 heap sort</font>
<font color="#FF0000">heap sort </font> ---- a fixed amount of additional storage
same time O(<img src="image/nlogn.gif" width="79" height="24" align="middle"> )
<i><font color="#0033CC">⊙slightly slower than merge sort using O(n) additional space
faster than merge sort using O(1) additional space</font></i>
void adjust<font color="#FF0000">(element list[ ] , int root, int n)
<i><font size="3" color="#CC0099">/* adj</font></i></font><i><font size="3" color="#CC0099">ust the binary tree to establish the heap */</font></i></font></b></pre>
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<pre><b><font face="Arial, Helvetica, sans-serif" size="4" color="#000000">{ int child,rootkey;
element temp;
temp = list[root];
rootkey = list[root].key;
child = 2*root; <i><font size="3" color="#CC0099">/* left child */</font></i>
while (child <= n){</font></b></pre>
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<pre align="left"><b><font face="Arial, Helvetica, sans-serif" size="4" color="#000000"> if ((child < n) && (list[child].key < list[child+1].key))
child++;
if (rootkey > list[child].key) /<i><font size="3" color="#CC0099">* compare root and max child */
</font></i> break;
else{
list[child/2] = list[child]; <i><font size="3" color="#CC0099"> /* move to parent */</font></i>
child *= 2;
}
}
list [child/2] = temp;
}
</font></b></pre>
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