以下代码可以从数组a[]中找出第k小的元素。它使用了类似快速排序中的分治算法,期望时间复杂度是O(N)的。请仔细阅读分析源码,填写划线部分缺失的内容。
#include
#include
int quick_select(int a[], int l, int r, int k) {
int p = rand() % (r - l + 1) + l;
int x = a[p];
{int t = a[p]; a[p] = a[r]; a[r] = t;}
int i = l, j = r;
while(i < j) {
while(i < j && a[i] < x) i++;
if(i < j) {
a[j] = a[i];
j--;
}
while(i < j && a[j] > x) j--;
if(i < j) {
a[i] = a[j];
i++;
}
}
a[i] = x;
p = i;
if(i - l + 1 == k) return a[i];
if(i - l + 1 < k) return quick_select( __________________ ); //填空
else return quick_select(a, l, i - 1, k);
}
int main()
{
int a[] = {1, 4, 2, 8, 5, 7, 23, 58, 16, 27, 55, 13, 26, 24, 12};
printf("%d\n", quick_select(a, 0, 14, 5));
return 0;
}
无
填空答案为:a,i+1,r,(k-i+l-1)
int main()
{
int a[] = {1, 4, 2, 8, 5, 7, 23, 58, 16, 27, 55, 13, 26, 24, 12};
printf("%d\n", quick_select(a, 0, 14, 5));
return 0;
}
把a数组赋值,并且传递四个参数;
int quick_select(int a[], int l, int r, int k) {
int p = rand() % (r - l + 1) + l;
int x = a[p];
{int t = a[p]; a[p] = a[r]; a[r] = t;}
int i = l, j = r;
由传递的四个参数我们可以知道:
l=0,r=14,k=5;
int p = rand() % (r - l + 1) + l;
//随机产生一个0~(r-l+1)的数,加上l后赋值给p;
int x = a[p];
//x赋值为a[p],也就是快排里面的基准-“主元”;
{int t = a[p]; a[p] = a[r]; a[r] = t;}
//把主元和最后一个元素交换位置;
int i = l, j = r;
i=0,j=14;
while(i < j) {
while(i < j && a[i] < x) i++;
if(i < j) {
a[j] = a[i];
j--;
}
while(i < j && a[j] > x) j--;
if(i < j) {
a[i] = a[j];
i++;
}
}
//标准的快速排序,小于主元的放前边,大于主元的放后面;
a[i] = x;
//把主元放到排序中的位置去;
p = i;
if(i - l + 1 == k) return a[i];
if(i - l + 1 < k)
return quick_select( __________________ ); //填空
else return quick_select(a, l, i - 1, k);
}
把i的位置用p记录下来,i-l-1就是i在这个数组中是第几位小的;
if(i - l + 1 == k) return a[i];
//假如第i-l+1位刚好和k相等,返回a[i];
if(i - l + 1 < k) return quick_select( __________________ ); //填空
//假如比k小,就应该在大的那边找;
else return quick_select(a, l, i - 1, k);
//假如比k大,就在小的那边找;
大的那边在右边,所以起始位置就应该:
l=i+1;而这样传过去,就相当于判断数组的左端点发生了改变,所以k在这里应该也要有相应的变化,即k=k-i+l-1;
所以填空的正确答案是
if(i - l + 1 < k)
return quick_select( a,i+1,r,(k-i+l-1) ); //填空
#include
#include
int quick_select(int a[], int l, int r, int k) {
int p = rand() % (r - l + 1) + l;
int x = a[p];
{int t = a[p]; a[p] = a[r]; a[r] = t;}
int i = l, j = r;
while(i < j) {
while(i < j && a[i] < x) i++;
if(i < j) {
a[j] = a[i];
j--;
}
while(i < j && a[j] > x) j--;
if(i < j) {
a[i] = a[j];
i++;
}
}
a[i] = x;
p = i;
if(i - l + 1 == k) return a[i];
if(i - l + 1 < k)
return quick_select( a,i+1,r,(k-i+l-1) ); //填空
else return quick_select(a, l, i - 1, k);
}
int main()
{
int a[] = {1, 4, 2, 8, 5, 7, 23, 58, 16, 27, 55, 13, 26, 24, 12};
printf("%d\n", quick_select(a, 0, 14, 5));
return 0;
}