算法小题目

一：求GCG（greatest common divisor)
analysis: tow nonnegative integer p and q, if q is 0 then gcd is q, then the gcd is the greatest common divisor of q and p%q. q和p%q的最大公约数，当然也是p和q的最大公约数。

int gcd(int p, int q){
  if(q == 0) return q;
  return gcd(q, p%q);
}

二：merge two sorted integer array into a third array, and make sure the third array also be sorted.
analysis: image two integer array are a and b, create array c and its length is
a.lenght plus b.length, define two varible indexA, indexB pointing the index of a and b, then loop infinite comparing a[indexA] and b[indexB], if a[indexA] > b[indexB], then assign b[indexB] to c[indexA+indexB] and indexB increase by 1, so vice verse. if  indexA exceeds a.length or indexB exceeds b.length, then copy the remaining of a or b to c and terminates the loop.
pseudo code:

int[] a = {};
int[] b = {};
int[] c = new int[a.length+b.length];
int indexA = indexB = 0;
while(true){
   while(true){
			if(indexA == a.length){// copy remaining b to c
				System.arraycopy(b, indexB, c, indexA+indexB, b.length - indexB);
				break;
			}
			if(indexB == b.length){//copy remaining a to c
				System.arraycopy(a, indexA, c, indexA+indexB, a.length-indexA);
				break;
			}
			
			if(a[indexA] > b[indexB]){
				c[indexA+indexB] = b[indexB];
				indexB++;
			}else{
				c[indexA+indexB] = a[indexA];
				indexA++;
			}
		}
}

三：reverse an array or testify if a number is a palindrome number
pseudo code:

int[] a = {};
int tmp;
for(int i=0;i<a.length/2;i++){
    tmp = a[i];
    a[i] = a[a.length-1-i];
    a[a.length-1-i] = tmp;
}

四：matrix-matrix multiplicatoin
在计算机中，一个矩阵说穿了就是一个二维数组。一个n行m列的矩阵可以乘以一个m行p列的矩阵，得到的结果是一个n行p列的矩阵，其中的第i行第j列位置上的数等于前一个矩阵第i行上的m个数与后一个矩阵第j列上的m个数对应相乘后所有m个乘积的和
pseudo code:

int[][] a = {{1, 1},{2, 0}};
		int[][] b = {{0, 2, 3}, {1, 1, 2}};
		if(a[0].length != b.length){
			throw new IllegalArgumentException("first matrix's column length must equals with second matrix's row length");
		}
		int[][] c = new int[a.length][b[0].length];
		for(int i=0;i<a.length;i++){//matrix a's row loop
			for(int j=0;j<b[0].length;j++){//matrix b's column loop
				for(int k=0;k<a[0].length;k++){
					c[i][j] += a[i][k] * b[k][j];
				}
			}
		}

五：HanoiTower
思想：首先要把前N-1个盘子转移到辅助柱，然后把第N个盘转到目标柱，最后再把前N-1个盘从辅助柱转移到目标柱，在转移过程中，把第三根柱子视为辅助柱。
pseudo code:

private static void hanoTower(String start, String dest, String auxiliary, int number){
		if(number == 0){
			return ;
		}
		hanoTower(start, auxiliary, dest, number-1);
		count++;
		System.out.println("move " + number + " from "+ start +" to "+ dest);
		hanoTower(auxiliary, dest, start, number-1);
	}

结论：总的移动次数是2^n -1， 如果N是偶数，则开始把1号盘移到辅助柱，如果N是奇数，把1号盘移到目标柱上。