兩道面試中的演算法題

給出隨機數列[-1,2,4,-5,0,2,4,6,9,-3,2],其最長單調遞增數列是[0,2,4,6,9]，寫出演算法。

 

#include <stdio.h>// 該演算法是我根據我草稿和答卷上的虛擬碼翻譯的 編譯運行看到的結果和我想的一樣。void get_longest_incr_array(int data[],int len) {int current_length = 1;int current_base = 0;int max_length = 1;int max_base = 0;for(int i = 1; i < len; i++ ) {if(data[i]>data[i-1]) {current_length += 1;if(current_length > max_length) {max_base = current_base;max_length = current_length;}} else {current_length = 1;current_base = i;}}printf("%d, %dn",max_base,max_length); // max_base是數組最長單調遞增子序列的起始點，max_length是該序列的長度};int main(int argc, char *argv[]){int test[14] = {5,4,1,2,3,4,5,6,1,2,5,3,2,1}; // 1,2,3,4,5,6get_longest_incr_array(test,14);int test2[14] = {0,2,1,2,1,1,3,5,7,9,10,11,12,14}; // 1,3,5,7,9,10,11,12,14get_longest_incr_array(test2,14);return 0;}

設有M*N矩陣

0,1,0,0,0,0,0,1,0,0
1,1,1,0,1,0,0,0,0,1
0,0,1,0,1,0,0,0,1,1
0,0,0,0,1,0,0,0,0,0
規則：有連續1的塊算一個大塊，例如上述矩陣有4個，求通用解法
【想了很久被面試官提示用遞迴尋找連續1塊，繼續想了很久得出最終解法  】

#include <stdio.h>int search_big_block(int *data,int m, int n,int index_x,int index_y) {if ( *(data+index_x*n+index_y) == 0 ){return 0;} else {*(data+index_x*n+index_y) = 0; // 順便置為0 如此遞迴下次尋找便不會找到這裡來了if(index_x > 0){search_big_block(data,m,n,index_x-1,index_y);}if(index_x < m){search_big_block(data,m,n,index_x+1,index_y);}if(index_y > 0){search_big_block(data,m,n,index_x,index_y-1);}if(index_y < n){search_big_block(data,m,n,index_x,index_y+1);}return 1;}}int main(int argc, char *argv[]){int data[] = {0,1,0,0,0,0,0,1,0,0,1,1,1,0,1,0,0,0,0,1,0,0,1,0,1,0,0,0,1,1,0,0,0,0,1,0,0,0,0,0};int control = 0;for(int i=0;i < 4;i++) {for(int j=0;j < 10;j++) {if (search_big_block(data,4,10,i,j)){control += 1;}}}printf("%d",control);return 0;}