• 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

    转载于:https://www.cnblogs.com/songdechiu/p/6611766.html

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  • 原文地址:https://www.cnblogs.com/twodog/p/12140977.html
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