复杂度分析 quick sort&merge sort

空间复杂度看新开了什么数据结构就够了

公式=几个点*每个点执行了多少次

二叉树都是n次

 

二分法查找：lgn

全部查找：n

n：找一个数，但是两边都要找。相当于遍历。类似于rotated sorted array的有重复 遍历版本。

nlgn：先分成两半，再全部合并。类似于merge sort.

//recursive and append
public static void mergeSort(int[] a, int n) {
    if (n < 2) {
        return;
    }
    int mid = n / 2;
    int[] l = new int[mid];
    int[] r = new int[n - mid];
 
    for (int i = 0; i < mid; i++) {
        l[i] = a[i];
    }
    for (int i = mid; i < n; i++) {
        r[i - mid] = a[i];
    }
    mergeSort(l, mid);
    mergeSort(r, n - mid);
 
    merge(a, l, r, mid, n - mid);
}

public static void merge(
  int[] a, int[] l, int[] r, int left, int right) {
  
    int i = 0, j = 0, k = 0;
    while (i < left && j < right) {
        if (l[i] < r[j]) {
            a[k++] = l[i++];
        }
        else {
            a[k++] = r[j++];
        }
    }
    while (i < left) {
        a[k++] = l[i++];
    }
    while (j < right) {
        a[k++] = r[j++];
    }
}

package com.java2novice.sorting;
 
public class MyQuickSort {
     
    private int array[];
    private int length;
 
    public void sort(int[] inputArr) {
         
        if (inputArr == null || inputArr.length == 0) {
            return;
        }
        this.array = inputArr;
        length = inputArr.length;
        quickSort(0, length - 1);
    }
 
    private void quickSort(int lowerIndex, int higherIndex) {
         
        int i = lowerIndex;
        int j = higherIndex;
        // calculate pivot number, I am taking pivot as middle index number
        int pivot = array[lowerIndex+(higherIndex-lowerIndex)/2];
        // Divide into two arrays
        while (i <= j) {
            /**
             * In each iteration, we will identify a number from left side which 
             * is greater then the pivot value, and also we will identify a number 
             * from right side which is less then the pivot value. Once the search 
             * is done, then we exchange both numbers.
             */
            while (array[i] < pivot) {
                i++;
            }
            while (array[j] > pivot) {
                j--;
            }
            if (i <= j) {
                exchangeNumbers(i, j);
                //move index to next position on both sides
                i++;
                j--;
            }
        }
        // call quickSort() method recursively
        if (lowerIndex < j)
            quickSort(lowerIndex, j);
        if (i < higherIndex)
            quickSort(i, higherIndex);
    }
 
    private void exchangeNumbers(int i, int j) {
        int temp = array[i];
        array[i] = array[j];
        array[j] = temp;
    }
     
    public static void main(String a[]){
         
        MyQuickSort sorter = new MyQuickSort();
        int[] input = {24,2,45,20,56,75,2,56,99,53,12};
        sorter.sort(input);
        for(int i:input){
            System.out.print(i);
            System.out.print(" ");
        }
    }
}

左边一直是最右的小数做索引，右边一直是最右的大数做索引。 

https://blog.csdn.net/github_30242787/article/details/50819414

动态规划约等于分治+记忆化

优点：快的原因：因为有了记忆化,所以算过的直接用就行,就不用再算一遍了.

缺点：dc属于recursion，要少用