There are now n ordered arrays in the M group, such as {1, 2, 3, 3}, {2, 3, 4, 6}, {1, 3, 5, 7}. In these arrays, select the data smaller than K, then return this value

Problem description: there are now n ordered arrays in M groups, such as {1, 2, 3, 4}, {2, 3, 6}, {1, 3, 5, 7 }, select the data smaller than K in these arrays and return this value.

Idea: Compare the minimum data selected each time by referring to the process of merging two Arrays

1. Define the selection position array index [m], initialized to 0

2. Find the Rochelle row Array Based on Index [m] each time to make sure that the current position of the Rochelle row array at the current time is the minimum value. Make sure that the value of index [I] is less than N.

3. Retrieve the minimum value. The index [l_row] is auto-incremented by 1.

4. Search until the k is found.

 1 public static int GetK_Min(int k) 2     { 3         int m=3,n=4; 4         if(k>m*n) 5             return -1; 6         int A[][] = {{1,3,5,6},{2,4,6,8},{1,2,4,6}}; 7          8         int index[] = {0,0,0}; 9         int l_min=0,l_row; 10         while(k>0)11         {12             for(int t=0;t<m;t++)13             {14                 if(index[t]<n)15                 {l_min =  A[t][index[t]];break;}16             }17             18             l_min = 100;19             l_row = 0;20             for(int i=0;i<m;i++)21             {22                 if(index[i]<n && l_min > A[i][index[i]])23                 {24                     l_min = A[i][index[i]];25                     l_row = i;26                 }27             }28             index[l_row] ++;29             30             k--;31         }32         33         return l_min;34     }

Algorithm analysis:

The time complexity of this algorithm is O (K * m), which is less complex than searching after merging and sorting.

There are now n ordered arrays in the M group, such as {1, 2, 3, 3}, {2, 3, 4, 6}, {1, 3, 5, 7}. In these arrays, select the data smaller than K, then return this value