[LintCode] Sort Integers II [Merge-sort, Quick-sort, Heap-sort]

linspiration

Problem

Given an integer array, sort it in ascending order. Use quick sort, merge sort, heap sort or any O(nlogn) algorithm.

Example

Given [3, 2, 1, 4, 5], return [1, 2, 3, 4, 5].

Note

考察对Heap Sort, Quick Sort, Merge Sort的掌握。

Solution

Merge Sort

public class Solution {
    public void sortIntegers2(int[] A) {
        if (A.length <= 1) return;
        int[] B = new int[A.length];
        sort(A, 0, A.length-1, B);
    }
    public void sort(int[] A, int start, int end, int[] B) {
        if (start >= end) return;
        int mid = start + (end - start) / 2;
        sort(A, start, mid, B);
        sort(A, mid+1, end, B);
        merge(A, start, mid, end, B);
    }
    public void merge(int[] A, int start, int mid, int end, int[] B) {
        int i = start, j = mid+1, index = start;
        while (i <= mid && right <= end) {
            if (A[i] < A[j]) B[index++] = A[i++];
            else B[index++] = A[j++];
        }
        while (i <= mid) B[index++] = A[i++];
        while (j <= end) B[index++] = A[j++];
        for (int k = start; k <= end; k++) A[k] = B[k];
    }
}

Heap Sort

public class Solution {
    private static int[] a;
    private static int n;
    private static int left;
    private static int right;
    private static int largest;
    public void sortIntegers2(int[] A) {
        a = A;
        buildheap(a);
        for(int i=n;i>0;i--){
            exchange(0, i);
            n=n-1;
            maxheap(a, 0);
        }
    }
    public static void buildheap(int []a){
        n=a.length-1;
        for(int i=n/2;i>=0;i--){
            maxheap(a,i);
        }
    }
    public static void maxheap(int[] a, int i){ 
        left=2*i;
        right=2*i+1;
        if(left <= n && a[left] > a[i]){
            largest=left;
        }
        else{
            largest=i;
        }
        
        if(right <= n && a[right] > a[largest]){
            largest=right;
        }
        if(largest!=i){
            exchange(i,largest);
            maxheap(a, largest);
        }
    }
    public static void exchange(int i, int j){
        int t=a[i];
        a[i]=a[j];
        a[j]=t; 
    }
}

Quick Sort

public class Solution {
    public void sortIntegers2(int[] A) {
        if (A.length <= 1) return;
        quicksort(A, 0, A.length-1);
    }
    public void quicksort(int[] A, int start, int end) {
        if (start >= end) return;
        int mid = start+(end-start)/2;
        int pivot = A[mid], i = start, j = end;
        while (i <= j) {
            while (i <= j && A[i] < pivot) i++;
            while (i <= j && A[j] > pivot) j--;
            if (i <= j) {
                int temp = A[i];
                A[i] = A[j];
                A[j] = temp;
                i++;
                j--;
            }
        }
        quicksort(A, start, j);
        quicksort(A, i, end);
    }
}
阅读 2.2k

Road to Glory
對酒當歌,人生幾何?譬如朝露,去日苦多。
161 声望
31 粉丝
0 条评论
你知道吗?

161 声望
31 粉丝
宣传栏