西安交通大学915-2019-编程3

题目描述：

解题思路：

• 暴力枚举，时间复杂度O(n^2)
• 归并排序，merge函数的拓展，一次处理一个区间的逆序对数量，时间复杂度O(nlogn)

代码实现：

• 暴力枚举

代码语言：javascript复制

#include <iostream>
#include <vector>
using namespace std;


int main() {
    vector<int> arr = {1, 3, 2, 5, 4, 7, 6};
    int n = arr.size();
    int cnt = 0;
    for (int i = 0; i < n; i  ) {
        for (int j = i   1; j < n; j  ) {
            // j = i   1  保证了i < j一定成立
            if (arr[i]  > arr[j]) {
                cnt  ;
            }
        }
    }
    cout << "the number of reverse number pair is " << cnt << endl;
    return 0;
}

• 最优算法

代码语言：javascript复制

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;


int cnt = 0; // 计数器，表示逆序对的个数

// left = [l, m-1]   right = [m, r]
void Merge(vector<int>& arr, int l, int m, int r) {
    int leftSize = m - l;
    int rightSize = r - m   1;
    int *left = new int[leftSize];
    int *right = new int[rightSize];
    for (int i = l; i < m; i  ) {
        left[i - l] = arr[i];
    }
    for (int i = m; i <= r; i  ) {
        right[i - m] = arr[i];
    }
    int i = 0;
    int j = 0;
    int k = l;
    while (i < leftSize && j < rightSize) {
        if (left[i] <= right[j]) {
            arr[k  ] = left[i  ];
        } else {
            arr[k  ] = right[j  ];
            // left[i] > right[j] && i < j 构成逆序
            // 此时左区间元素大于右区间元素，则左区间的所有剩余元素都与右区间的当前元素构成逆序对 
            cnt  = (leftSize - i);
        }
    }
    while (i < leftSize) {
        arr[k  ] = left[i  ];
    }

    while (j < rightSize) {
        arr[k  ] = right[j  ];
    }

    delete[] left;
    left = nullptr;
    delete[] right;
    right = nullptr;
}
void _MergeSort(vector<int> &arr, int l, int r) {
    if (l >= r) {
        return;
    }
    int m = (r - l) / 2   l;
    _MergeSort(arr, l, m);
    _MergeSort(arr, m   1, r);
    Merge(arr, l, m   1, r);
}

void MergeSort(vector<int> &arr) {
    _MergeSort(arr, 0, arr.size() - 1);
}

int main() {
    vector<int> arr = {1, 3, 2, 5, 4, 7, 6};
    auto f = [] (int x) {cout << x << " ";};
    for_each(arr.begin(), arr.end(), f); cout << endl;
    MergeSort(arr);
    for_each(arr.begin(), arr.end(), f); cout << endl;
    cout << "the number of reverse number pair is " << cnt << endl;
    return 0;
}

merge 函数 排序 算法

0 人点赞