算法（Algorithms）第4版 练习 2.2.10

关键代码实现：

private static void merge(Comparable[] input, int lo, int mid, int hi) {

//copy input[lo,mid] to aux[lo,mid]
for(int i = lo; i <= mid; i++) {
aux[i] = input[i];
}
//copy input[mid+1,hi] to aux[hi,mid+1]
for(int i = mid+1; i <= hi; i++) {
aux[i] = input[hi-i+mid+1];
}

int i = lo;
int j = hi;
for(int k = lo; k <= hi; k++) {
if(less(aux[j], aux[i]))
input[k] = aux[j--];
else
input[k] = aux[i++];
}

StdOut.printf("merge(input, %4d, %4d, %4d)", lo, mid, hi);
show(input);//for test

}

分析：

当j从hi减到mid的时候，less(aux[j],aux[i])永为false，则一直执行input[k] = aux[i++]。

当i从lo增至mid+1时，less(aux[j],aux[i])永为true，则一直执行input[k] = aux[j--]。

所以可以移除索引i和j出界判断的代码。

测试结果：

M E R G E S O R T E X A M P L E
merge(input,    0,    0,    1)E M R G E S O R T E X A M P L E
merge(input,    2,    2,    3)E M G R E S O R T E X A M P L E
merge(input,    0,    1,    3)E G M R E S O R T E X A M P L E
merge(input,    4,    4,    5)E G M R E S O R T E X A M P L E
merge(input,    6,    6,    7)E G M R E S O R T E X A M P L E
merge(input,    4,    5,    7)E G M R E O R S T E X A M P L E
merge(input,    0,    3,    7)E E G M O R R S T E X A M P L E
merge(input,    8,    8,    9)E E G M O R R S E T X A M P L E
merge(input,   10,   10,   11)E E G M O R R S E T A X M P L E
merge(input,    8,    9,   11)E E G M O R R S A E T X M P L E
merge(input,   12,   12,   13)E E G M O R R S A E T X M P L E
merge(input,   14,   14,   15)E E G M O R R S A E T X M P E L
merge(input,   12,   13,   15)E E G M O R R S A E T X E L M P
merge(input,    8,   11,   15)E E G M O R R S A E E L M P T X
merge(input,    0,    7,   15)A E E E E G L M M O P R R S T X
A E E E E G L M M O P R R S T X

速度对比：

For 20000 random Doubles 1000 trials
Merge is 3.4s FasterMerge is 3.1s