子数组换位问题 C语言

#include <stdio.h>
//设a[0:n-1]是一个有n个元素的数组，k(0<=k<=n-1)是一个非负整数。 
//试设计一个算法将子数组a[0:k-1]与a[k,n-1]换位。要求算法在最坏情况下耗时O(n),且只用到O(1)的辅助空间。
void swap(int a[],int low1, int high1, int low2) {
	int temp;
	while(low1 <= high1) {
		temp = a[low1];
		a[low1] = a[low2];
		a[low2] = temp;
		low1++;
		low2++;
	}
}
void pation(int a[],int low, int high, int k) {
	if(low < high) {
		if((k-low) == (high-k+1)) {//等长直接交换
			swap(a,low,k-1,k);
		}else if((k-low) > (high-k+1)) {//前段大于后段，将前段与后段等长交换
			swap(a,low,low+high-k,k);
			pation(a,low+high-k+1,high,k);
		}else{//后段大于前段，将后段与前段等长交换
			swap(a,low,k-1,high+low-k+1);
			pation(a,low,high+low-k,k); 
		}
	}
	return;
}
int main() {
	int n;
	scanf("%d",&n);
	int a[n],k;
	for(int i = 0;i < n; i++) {
		scanf("%d",&a[i]);
	}
	scanf("%d",&k);
	pation(a,0,n-1,k);
	for(int j = 0; j < n; j++) {
		printf("%d ",a[j]);
	}
}

标签：前段,int,换位,scanf,C语言,high,low,数组,pation
来源： https://blog.csdn.net/qq_41841130/article/details/94049057