编程之美2.10之寻找数组中的最大值和最小值

问题描述：寻找一个数组中的最大值和最小值。
方法一：扫描一遍数组，找出最大和最小

方法二：将相邻两个数分为一组，较大的数放在偶数位、较小的数放在奇数位。再分别从这两部分中找出最大值和最小值。

方法三：将相邻两个数分为一组，用Max和Min变量存储当前的最大值和最小值。同一数组比较之后，不交换次序，而是将下一个两个数的较大值和Max比较，较小数和Min比较。以此往后执行。

方法四：采用分治思想，分别求前后N/2个数的Min,Max 。然后取较小的Min和较大的Max

package suda.alex.chapter2;
import java.util.*;
public class SearchMaxMin {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner scanner = new Scanner(System.in);
        System.out.println("please input the size of array:");
        int n = scanner.nextInt();
        System.out.println("please input an array:");
        int a[] = new int[n];
        for(int i = 0;i < n;i++){
            a[i] = scanner.nextInt();
        }
// System.out.println("Max is:" + searchMax(a));
// System.out.println("Min is:" + searchMin(a));
        System.out.println("Max is:" + searchMaxMin3(a,0,n-1)[0]);
        System.out.println("Min is:" + searchMaxMin3(a,0,n-1)[1]);

    }
    public static int searchMax(int[] a){
        int max = a[0];
        for(int i = 1;i < a.length;i++){
            if(a[i] > max){
                max = a[i];
            }
        }
        return max;
    }
    public static int searchMin(int[] a){
        int min = a[0];
        for(int i = 1;i < a.length;i++){
            if(a[i] < min){
                min = a[i];
            }
        }
        return min;
    }
    public static int[] searchMaxMin1(int[] a){
        int len = a.length;
        int[] even;
        int[] odd = new int[len/2];
        if(len%2 == 0){
            even = new int[len/2];
            int count = 0;
            for(int i = 0;i < len-1;i+=2){
                if(a[i] < a[i+1]){
                    int t = a[i];
                    a[i] = a[i+1];
                    a[i+1] = t;
                }
                even[count] = a[i];
                odd[count] = a[i+1];
                count++;
            }

        }
        else{
            even = new int[len/2 + 1];
            int count = 0;
            for(int i = 0;i < len-2;i+=2){
                if(a[i] < a[i+1]){
                    int t = a[i];
                    a[i] = a[i+1];
                    a[i+1] = t;
                }
                even[count] = a[i];
                odd[count] = a[i+1];
                count++;
            }
            even[len/2] = a[len-1];
        }
        int max = searchMax(even);
        int min = searchMin(odd);
        int[] result = {max,min};
        return result;
    }
    public static int[] searchMaxMin2(int[] a){
        int len = a.length;
        int min,max;
        if(a[0] < a[1]){
            min = a[0];
            max = a[1];
        }
        else{
            min = a[1];
            max = a[0];
        }
        if (len % 2 == 0) {
            for (int i = 2; i < len; i += 2) {
                if (a[i] > a[i + 1]) {
                    if (a[i] > max) {
                        max = a[i];
                    }
                    if (a[i + 1] < min) {
                        min = a[i + 1];
                    }
                } else {
                    if (a[i + 1] > max) {
                        max = a[i + 1];
                    }
                    if (a[i] < min) {
                        min = a[i];
                    }
                }

            }
        }
        else {
            for (int i = 2; i < len-2; i += 2) {
                if (a[i] > a[i + 1]) {
                    if (a[i] > max) {
                        max = a[i];
                    }
                    if (a[i + 1] < min) {
                        min = a[i + 1];
                    }
                } else {
                    if (a[i + 1] > max) {
                        max = a[i + 1];
                    }
                    if (a[i] < min) {
                        min = a[i];
                    }
                }
                if(a[len-1] < min){
                    min = a[len-1];
                }
                if(a[len-1] > max){
                    max = a[len-1];
                }

            }
        }
        int[] results = {max,min};
        return results;

    }
    //分治思想 分别求前后N/2个数的Min,Max 然后去较小的Min和较大的Max
    public static int[] searchMaxMin3(int[] a, int low, int high) {
        int[] result = new int[2];
        if (high - low <= 1) {
            if (a[low] < a[high]) {
                result[0] = a[high];
                result[1] = a[low];
                return result;
            } else {
                result[0] = a[low];
                result[1] = a[high];
                return result;
            }

        }
        int[] resultLeft = searchMaxMin3(a, low, low + (high-low)/2);
        int[] resultRight = searchMaxMin3(a, low + (high-low)/2, high);
        if(resultLeft[0] > resultRight[0]){
            result[0] = resultLeft[0];
        }
        else{
            result[0] = resultRight[0];
        }
        if(resultLeft[1] < resultRight[1]){
            result[1] = resultLeft[1];
        }
        else{
            result[1] = resultRight[1];
        }
        return result;


    }

}