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51_InversePairs.cpp
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174 lines (136 loc) · 3.6 KB
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/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题51：数组中的逆序对
// 题目：在数组中的两个数字如果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。
#include <cstdio>
int InversePairsCore(int* data, int* copy, int start, int end);
int InversePairs(int* data, int length)
{
if(data == nullptr || 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 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);
// i初始化为前半段最后一个数字的下标
int i = start + length;
// j初始化为后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;
}
// ====================测试代码====================
void Test(char* testName, int* data, int length, int expected)
{
if(testName != nullptr)
printf("%s begins: ", testName);
if(InversePairs(data, length) == expected)
printf("Passed.\n");
else
printf("Failed.\n");
}
void Test1()
{
int data[] = { 1, 2, 3, 4, 7, 6, 5 };
int expected = 3;
Test("Test1", data, sizeof(data) / sizeof(int), expected);
}
// 递减排序数组
void Test2()
{
int data[] = { 6, 5, 4, 3, 2, 1 };
int expected = 15;
Test("Test2", data, sizeof(data) / sizeof(int), expected);
}
// 递增排序数组
void Test3()
{
int data[] = { 1, 2, 3, 4, 5, 6 };
int expected = 0;
Test("Test3", data, sizeof(data) / sizeof(int), expected);
}
// 数组中只有一个数字
void Test4()
{
int data[] = { 1 };
int expected = 0;
Test("Test4", data, sizeof(data) / sizeof(int), expected);
}
// 数组中只有两个数字，递增排序
void Test5()
{
int data[] = { 1, 2 };
int expected = 0;
Test("Test5", data, sizeof(data) / sizeof(int), expected);
}
// 数组中只有两个数字，递减排序
void Test6()
{
int data[] = { 2, 1 };
int expected = 1;
Test("Test6", data, sizeof(data) / sizeof(int), expected);
}
// 数组中有相等的数字
void Test7()
{
int data[] = { 1, 2, 1, 2, 1 };
int expected = 3;
Test("Test7", data, sizeof(data) / sizeof(int), expected);
}
void Test8()
{
int expected = 0;
Test("Test8", nullptr, 0, expected);
}
int main(int argc, char* argv[])
{
Test1();
Test2();
Test3();
Test4();
Test5();
Test6();
Test7();
Test8();
return 0;
}