编程算法

数组中的逆序对 代码(C)

本文地址: http://blog.csdn.net/caroline_wendy

题目: 在数组中的两个数字假设前面一个数字大于后面的数字, 则这两个数字组成一个逆序对.

输入一个数组, 求出这个数组中的逆序对的总数.

使用归并排序的方法, 辅助空间一个排序的数组, 依次比較前面较大的数字, 算出总体的逆序对数, 不用逐个比較.

时间复杂度: O(nlogn)

代码:

/*
 * main.cpp
 *
 *  Created on: 2014.6.12
 *      Author: Spike
 */

/*eclipse cdt, gcc 4.8.1*/

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int InversePairsCore(int* data, int* copy, int start, int end) {
	if (start == end) {
		copy[start] = data[start];
		return 0;
	}
	int length = (end-start)/2;
	int left = InversePairsCore(copy, data, start, start+length);
	int right = InversePairsCore(copy, data, start+length+1, end);

	int i = start+length; //前半段最后一个数字的下标
	int j = end;
	int indexCopy = end;
	int count = 0;
	while (i>=start && j>=start+length+1) {
		if (data[i] > data[j]) {
			copy[indexCopy--] = data[i--];
			count += j-start-length;
		} else {
			copy[indexCopy--] = data[j--];
		}
	}
	for (; i>=start; --i)
		copy[indexCopy--] = data[i];
	for (; j>=start+length+1; --j)
		copy[indexCopy--] = data[j];
	return left+right+count;

}

int InversePairs (int* data, int length) {
	if (data == NULL || length < 0)
		return 0;
	int *copy = new int[length];
	for (int i=0; i<length; ++i)
		copy[i] = data[i];
	int count = InversePairsCore(data, copy, 0, length-1);
	delete[] copy;
	return count;
}


int main(void)
{
	int data[] = {7, 5, 6, 4};
	int result =  InversePairs (data, 4);
    printf("result = %d
", result);

    return 0;
}

输出:

result = 5