[LC总结] 排序 Median [QuickSort] Sort Integers II

opengps 发布于2019-08-14 15:35 / 2347人阅读

Problem

Given a binary search tree and a new tree node, insert the node into the tree. You should keep the tree still be a valid binary search tree.
Given a unsorted array with integers, find the median of it.

A median is the middle number of the array after it is sorted.

If there are even numbers in the array, return the N/2-th number after sorted.

Example

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

Given [7, 9, 4, 5], return 5.

Challenge

O(n) time.

Note

理解快排。注意，作为pivot的元素在递归时要exclude出来。

Solution

Last as pivot

public class Solution {
    public int median(int[] nums) {
        return helper(nums, 0, nums.length-1, (nums.length-1)/2);
    }
    public int helper(int[] A, int start, int end, int k) {
        int l = start, r = end;
        int pivot = end, a = A[pivot];
        while (l < r) {
            while (l < r && A[l] < a) l++;
            while (l < r && A[r] >= a) r--;
            swap(A, l, r);
        }
        swap(A, l, end);
        if (l == k) return A[l];
        else if (l < k) return helper(A, l+1, end, k);
        else return helper(A, start, l-1, k);
    }
    public void swap(int[] A, int l, int r) {
        int temp = A[l];
        A[l] = A[r];
        A[r] = temp;
    }
}

Sort Integers II Merge sort

public class Solution {
    /**
     * @param A an integer array
     * @return void
     */
    public void sortIntegers2(int[] A) {
        // Write your code here
        if (A.length <= 1) return;
        int[] B = new int[A.length];
        sort(A, 0, A.length-1, B);
    }
    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);
    }
    void merge(int[] A, int start, int mid, int end, int[] B) {
        int i = start, j = mid+1, index = start;
        while (i <= mid && j <= end) {
            if (A[i] < A[j]) B[index++] = A[i++];
            else B[index++] = A[j++];
        }
        while (j <= end) B[index++] = A[j++];
        while (i <= mid) B[index++] = A[i++];

        for (int k = start; k <= end; k++) A[k] = B[k];
    }
}

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 i = start, j = end;
        int pivot = A[start+(end-start)/2];
        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);
    }
}

Heap Sort

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

GPU云服务器云服务器 sort排序算法排序算法总结 Median Integers

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