编程之美2.10之寻找数组中的最大值和最小值

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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;    }}

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