算法——归并排序

public class Merge {

    private static Comparable[] aue;

    public static void sort(Comparable[] a){
        aue = new Comparable[a.length];
        sort(a,0,a.length-1);
    }

    public static void sort(Comparable[] a,int lo,int hi){
        if(hi <= lo ) return;
        int mid = (lo + hi)/2;
        sort(a,lo,mid);
        sort(a,mid+1,hi);
        merge(a,lo,mid,hi);
    }

    public static void merge(Comparable[] a,int lo,int mid,int hi){
        int i = lo;
        int j = mid + 1;
        for(int k = lo; k <= hi;k++){
            aue[k] = a[k];
        }

        for(int k = lo; k <= hi;k++){
            if( i > mid) a[k] = aue[j++];
            else if( j > hi) a[k] = aue[i++];
            else if(less(aue[j],aue[i])) a[k] = aue[j++];
            else a[k] = aue[i++];
        }
    }


    public static boolean less(Comparable v,Comparable w){
        return v.compareTo(w) < 0;
    }

    public static void echo(Comparable[] a,int i,int j){
        Comparable t = a[i];
        a[i] = a[j];
        a[j] = t;
    }

    public static void show(Comparable[] a){
        for(int i = 0; i < a.length;i++ ){
            System.out.print(a[i] + " ");
        }
    }

    public static boolean isSorted(Comparable[] a){
        for(int i = 1; i < a.length;i++){
            if(less(a[i],a[i-1])) return false;
        }
        return true;
    }

    public static Comparable[] Random(int num){
        Integer[] a = new Integer[num];
        Random rand = new Random();
        for(int i = 0;i<num;i++){
            a[i] =  rand.nextInt(100000);
        }
        return a;
    }

    public static void main(String[] args) {
        Comparable[] a = Selection.Random(2000000);
        /*show(a);*/
        Long start = System.currentTimeMillis();
        sort(a);
        Long end = System.currentTimeMillis();
        System.out.println("Time is " + (end - start));
        /*show(a);*/
    }
}

思路：利用分治思想，把大问题拆分为小问题，然后通过小问题的答案来解决大问题的答案， 归并重点是分为两步，一个是拆分，通过递归不断的对半分，直到划分为一个数时就返回，另一个是合并，通过增加一个额外的数组,长度为N,利用循环进行4个判断，左半部分用尽就用右半部，反之亦然，右半边小于左半边的情况和右半边大于左半边的情况，这里要注意j=mid + 1,一定要加1，这样mid会往右边偏移，才不会出错。