矩阵原地转置，空间复杂度为O(1)(暂时先保存下来)

#include<iostream>using namespace std; /* 后继 */int getNext(int i, int m, int n){    return (i%n)*m + i/n;} /* 前驱 */int getPre(int i, int m, int n){    return (i%m)*n + i/m;} /* 处理以下标i为起点的环 */void movedata(int *mtx, int i, int m, int n){    int temp = mtx[i];  // 暂存    int cur = i;       // 当前下标    int pre = getPre(cur, m, n);    while(pre != i)    {        mtx[cur] = mtx[pre];        cur = pre;        pre = getPre(cur, m, n);    }    mtx[cur] = temp;} /* 转置，即循环处理所有环 */void transpose(int *mtx, int m, int n){    for(int i=0; i<m*n; ++i)    {        int next = getNext(i, m, n);        while(next > i) // 若存在后继小于i说明重复            next = getNext(next, m, n);        if(next == i)   // 处理当前环             movedata(mtx, i, m, n);    }} /* 输出矩阵 */void print(int *mtx, int m, int n){    for(int i=0; i<m*n; ++i)    {        if((i+1)%n == 0)            cout << mtx[i] << "\n";        else            cout << mtx[i] << " ";    }} /* 测试 */int main(){    int matrix[4*2] = {1,2,3,4,5,6,7,8};    cout << "Before matrix transposition:" << endl;    print(matrix, 4, 2);    transpose(matrix, 4, 2);    cout << "After matrix transposition:" << endl;    print(matrix, 2, 4);    return 0;}