Merge Sorted Array

描述
Given two sorted integer arrays A and B, merge B into A as one sorted array.
Note: You may assume that A has enough space to hold additional elements from B. The number of
elements initialized in A and B are m and n respectively.

代码

package sort;

public class MergeSortedArray {
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		// 这道题是说让B merge到 A 里面。
		int[] A = { 1, 14, 15, 19 };
		int m=4;
		int n=6;
		int[] B = { 2, 3, 17, 20, 34, 67 };
		int[] C = mergeTwoList(A, B);
		for (int k : C) {
			System.out.println(k);
		}
		ListNode left = new ListNode(1);
		// head.add(1);
		left.add(3);
		left.add(7);
		left.add(11);
		left.add(13);
		left.add(15);
		left.print();
		System.out.println();
		ListNode right = new ListNode(2);
		// head.add(1);
		right.add(4);
		right.add(6);
		right.add(8);
		right.add(10);
		right.add(16);
		right.print();
		System.out.println();
		ListNode clistnode = mergeTwoList(left, right);
		clistnode.print();
		
		merge(A,m,B,n);
		for(int p:A) {
			System.out.println(p);
		}
	}

	public static int[] mergeTwoList(int[] A, int[] B) {
		int[] C = new int[A.length + B.length];
		int k = 0;
		int i = 0;
		int j = 0;
		while (i < A.length && j < B.length) {
			if (A[i] < B[j])
				C[k++] = A[i++];
			else
				C[k++] = B[j++];
		}

		while (i < A.length) {
			C[k++] = A[i++];
		}
		while (j < B.length) {
			C[k++] = B[j++];
		}
		return C;
	}

	public static ListNode mergeTwoList(ListNode leftlist,ListNode rightlist) {
		if(rightlist==null) {
			return leftlist;
		}
		if(leftlist==null) {
			return rightlist;
		}
		ListNode fakehead=new ListNode(-1);
		ListNode ptr=fakehead;
		while(rightlist!=null&&leftlist!=null) {
			if(rightlist.data<leftlist.data) {
				ptr.next=rightlist;
				ptr=ptr.next;
				rightlist=rightlist.next;
			}else {
				ptr.next=leftlist;
				ptr=ptr.next;
				leftlist=leftlist.next;
			}
		}
		if(rightlist!=null) {
			ptr.next=rightlist;
		}
		if(leftlist!=null) {
			ptr.next=leftlist;
		}
		
		return fakehead.next;
	}

	// 这道题是不能借助一个新的array的,那么我们就不好从前往后比了(不好插入位置)。
	// 方便的方法是从后往前比,然后最后处理剩下的元素。
	public static void merge(int[] A,int m,int[] B,int n) {
		while(m>0&&n>0) {
			if(A[m-1]>B[n-1]) {
				A[m+n-1]=A[m-1];
				m--;
			}else {
				A[m+n-1]=B[n-1];
				n--;
			}
		}
		while(n>0) {
			A[m+n-1]=B[n-1];
			n--;
		}
	}
	
}

  归并排序算法

将待排序序列R[0...n-1]看成是n个长度为1的有序序列,将相邻的有序表成对归并,得到n/2个长度为2的有序表;将这些有序序列再次归并,得到n/4个长度为4的有序序列;如此反复进行下去,最后得到一个长度为n的有序序列。

综上可知:

归并排序其实要做两件事:

(1)“分解”——将序列每次折半划分

(2)“合并”——将划分后的序列段两两合并后排序

package sort;

public class MergeSort {

    public static void main(String[] args) {
        // TODO Auto-generated method stub

        int[] array = { 9, 1, 5, 3, 4, 2, 6, 8, 7 };

        MergeSort merge = new MergeSort();
        System.out.print("排序前:		");
        merge.printAll(array);
        merge.sort(array);
        System.out.print("排序后:		");
        merge.printAll(array);

    }

    public void Merge(int[] array, int low, int mid, int high) {
        int i = low; // i是第一段序列的下标
        int j = mid + 1; // j是第二段序列的下标
        int k = 0; // k是临时存放合并序列的下标
        int[] array2 = new int[high - low + 1]; // array2是临时合并序列

        // 扫描第一段和第二段序列,直到有一个扫描结束
        while (i <= mid && j <= high) {
            // 判断第一段和第二段取出的数哪个更小,将其存入合并序列,并继续向下扫描
            if (array[i] <= array[j]) {
                array2[k] = array[i];
                i++;
                k++;
            } else {
                array2[k] = array[j];
                j++;
                k++;
            }
        }

        // 若第一段序列还没扫描完,将其全部复制到合并序列
        while (i <= mid) {
            array2[k] = array[i];
            i++;
            k++;
        }

        // 若第二段序列还没扫描完,将其全部复制到合并序列
        while (j <= high) {
            array2[k] = array[j];
            j++;
            k++;
        }

        // 将合并序列复制到原始序列中
        for (k = 0, i = low; i <= high; i++, k++) {
            array[i] = array2[k];
        }
    }

    public void MergePass(int[] array, int gap, int length) {
        int i = 0;

        // 归并gap长度的两个相邻子表
        for (i = 0; i + 2 * gap - 1 < length; i = i + 2 * gap) {
            Merge(array, i, i + gap - 1, i + 2 * gap - 1);
        }

        // 余下两个子表,后者长度小于gap
        if (i + gap - 1 < length) {
            Merge(array, i, i + gap - 1, length - 1);
        }
    }

    public int[] sort(int[] list) {
        for (int gap = 1; gap < list.length; gap = 2 * gap) {
            MergePass(list, gap, list.length);
            System.out.print("gap = " + gap + ":	");
            this.printAll(list);
        }
        return list;
    }

    public void printAll(int[] list) {
        for (int value : list) {
            System.out.print(value + "	");
        }
        System.out.println();
    }
}
原文地址:https://www.cnblogs.com/ncznx/p/9590340.html