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算逆序数可以O(n)完成的int count_inver(char *str, int len)
{
int i;
int cnt = 0;
int a[4] = {0};
for(i = len - 1; i >= 0; i--) {
switch (str[i]) {
case 'A':
a[1]++;
a[2]++;
a[3]++;
break;
case 'C':
a[2]++;
a[3]++;
cnt += a[1];
break;
case 'G':
a[3]++;
cnt += a[2];
break;
case 'T':
cnt += a[3];
}
}
return cnt;
}
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