最大子序列问题

时间：2014-08-03 17:59:35 阅读：263 评论：0 收藏：0 [点我收藏+]

最大子数组问题

定义

给定整数A₁, A_2, …, A_n(其中可能是负数)，求_k的最大值和序列的起始位置（为了方便起见，如果所有整数均为负数，则最大子序列和为0），使用四种算法（根据运行时间区分）解决这个问题。

运行时间为θ(n³)

使用了三个for循环，在最坏情况下，运行时间为θ(n³)

C语言实现代码

#include<stdio.h>
#define LEN(array) (sizeof(array)/sizeof(array[0])) 

void maxsubarray(int *array, int len, int *, int *);

main(){
    int array[] = {13, -3, -25, 20, -3, -16, -23, 18, 20, -7, 12, -5, -22, 15, -4, 7};
    int i, start, end;
    
    printf("原始数组:\n");
    for (i = 0; i < LEN(array); i++)
        printf("%5d", array[i]);
        
    putchar(‘\n‘);
    maxsubarray(array, LEN(array), &start, &end);
    printf("最大子序列的起始位置为：%d, 终止位置为：%d\n", start+1, end+1);
    printf("最大子序列为: ");    
    
    for (i = start; i <= end; i++)
        printf("%5d", array[i]);
    putchar(‘\n‘);
    putchar(‘\n‘);
}

void maxsubarray(int *array, int len, int *start, int *end){
    int i, j, k, sum, maxsum;
    
    maxsum = 0;
    for (i = 0; i < len; i++){
        for (j = i; j < len; j++){
            sum = 0;
            for (k = i; k <= j; k++){
                sum += array[k];
                if (sum > maxsum){
                    maxsum = sum;
                    *start = i;
                    *end = j;
                }                    
            }
        }
    }
    
    printf("\n子序列的求和最大值为： %d\n", maxsum);        
}

运行时间为θ(n²)

使用了两个for循环，最坏情况下运行时间为θ(n²)

C语言实现代码

#include<stdio.h>

#define LEN(array) (sizeof(array)/sizeof(array[0]))

 

void maxsubarray(int *array, int len, int *, int *);

 

main(){

         int array[] = {13, -3, -25, 20, -3, -16, -23, 18, 20, -7, 12, -5, -22, 15, -4, 7};

         int i, start, end;

        

         printf("原始数组:\n");

         for (i = 0; i < LEN(array); i++)

                   printf("%5d", array[i]);

                  

         putchar(‘\n‘);

         maxsubarray(array, LEN(array), &start, &end);

         printf("最大子序列的起始位置为：%d, 终止位置为：%d\n", start+1, end+1);

         printf("最大子序列为: ");

        

         for (i = start; i <= end; i++)

                   printf("%5d", array[i]);

         putchar(‘\n‘);

         putchar(‘\n‘);

}

 

void maxsubarray(int *array, int len, int *start, int *end){

         int i, j, sum, maxsum;

        

         maxsum = 0;

         for (i = 0; i < len; i++){

                   sum = 0;

                   for (j = i; j < len; j++){                          

                                     sum += array[j];

                                     if (sum > maxsum){

                                               maxsum = sum;

                                               *start = i;

                                               *end = j;

                                     }                                            

                           

                   }

         }

        

         printf("\n子序列的求和最大值为： %d\n", maxsum);             

}

运行时间为θ(nlgn)

使用分治法求解：

假设我们要寻找子数组A[low..high]的最大子数组，使用分治技术意味着要将子数组划分为两个规模尽量相等的子数组，也就是说，要找到子数组的中央位置mid，然后考虑求解两个子数组A[low..high]的任何连续子数组A[i..j]所处的位置必然是以下三种情况之一：

完全位于子数组A[low, mid]中，因此，low <= i <= j <= mid
完全位于子数组A[mid+1, high]中，因此mid <= i <= j <= high
跨越了重点，因此low <= i <= mid < j <= high

递归地求解A[low..mid]和A[mid+1..high]的最大子数组，因为这两个子问题仍然是最大子数组问题，只是规模更小

伪代码

注意求出跨越中点的最大子数组问题并非原问题的更小实例，因为它加入了限制，求出的子数组必须跨越中点，任何跨越中点的子数组都由两个子数组A[i..mid]和A[mid+1..j]组成，其中low<= i <= mid且mid< j <= high，因此，只需要找出形如A[i..mid]和A[mid+1..j]的最大子数组，然后将其合并即可，过程FIND-MAX-CORSSING-SUBARRAY接收数组A和下标low, mid和high为输入，返回一个下标元组划定跨越中点的最大子数组的边界，并返回最大子数组中值的和

先求出左半部A[low..mid]的最大子数组，再求出右半部A[mid+1..high]的最大子数组，然后返回子数组A[max-left..max-right]

  FIND-MAX-CORSSING-SUBARRAY(A, low, mid, high)

         left-sum = A[mid]

         sum = 0

         for i = mid downto low

                   sum = sum + A[i]

                   if sum > left-sum

                            left-sum = sum

                            max-left = i

         right-sum = A[mid+1]

         sum = 0

         for j = mid+1 to high

                   sum += A[j]

                   if sum > right-sum

                            right-sum = sum

                            max-right = j

         return (max-left, max-right, left-sum+right-sum)

求解最大子数组问题的分治算法的伪代码：

FIND-MAXIMUM-SUBARRAY(A, low, high)

         if high == low

                   return(low, high, A[low])

         else mid = (low + high) / 2

                   (left-low, left-high, left-sum) = FIND-MAXIMUM-SUBARRAY(A, low, mid)

                   (right-low, right-high, right-sum) = FIND-MAXIMUM-SUBARRAY(A, mid+1, high)

                   (cross-low, corss-high, cross-sum) = FIND-MAX-CROSSING-SUBARRAY(A, low, mid, high)

                   if left-sum >= right-sum and left-sum >= corss-sum

                            return (left-low, left-high, left-sum)

                   else if right-sum >= left-sum and right-sum >= cross-sum

                            return (right-low, right-high, right-sum)

                   else

                            return (corss-low, cross-high, cross-sum)

C语言代码实现

#include<stdio.h>

#define LEN(array) (sizeof(array)/sizeof(array[0]))

 

int maxsubarray(int *, int , int , int *, int *);

int crosssubarray(int *array, int low, int mid, int high, int *begin, int *end);

 

main(){

         int array[] = {13, -3, -25, 20, -3, -16, 23, -18, -20, -7, 12, -5, -22, 15, -4};

         int i, start, end, sum;

        

         printf("原始数组:\n");

         for (i = 0; i < LEN(array); i++)

                   printf("%5d", array[i]);

                  

         putchar(‘\n‘);

         sum = maxsubarray(array, 0, LEN(array)-1, &start, &end);

         printf("最大子序列的和为：%d\n", sum);

         printf("最大子序列的起始位置为：%d, 终止位置为：%d\n", start+1, end+1);

         printf("最大子序列为: ");

        

         for (i = start; i <= end; i++)

                   printf("%5d", array[i]);

         putchar(‘\n‘);

         putchar(‘\n‘);

}

 

int maxsubarray(int *array, int low, int high, int *begin, int *end){

         int mid, left_sum, right_sum, cross_sum;

         int left_low, left_high, right_low, right_high, cross_low, cross_high;

        

         if (low == high){

                   return array[low];

                   *begin = low;

                   *end = high;

         }

                  

         mid = (low + high) / 2;

         left_sum = maxsubarray(array, low, mid, begin, end);

         left_low = *begin;

         left_high = *end;

         right_sum = maxsubarray(array, mid+1, high, begin, end);

         right_low = *begin;

         right_high = *end;

         cross_sum = crosssubarray(array, low, mid, high, begin, end);

         cross_low = *begin;

         cross_high = *end;

        

         if (left_sum > right_sum && left_sum > cross_sum){

                   *begin = left_low;

                   *end = left_high;

                   return left_sum;

         }

                  

         else if (right_sum > left_sum && right_sum > cross_sum){

                   *begin = right_low;

                   *end = right_high;

                   return right_sum;

         }                

         else{

                   *begin = cross_low;

                   *end = cross_high;

                   return cross_sum;             

         }

                           

}

 

int crosssubarray(int *array, int low, int mid, int high, int *begin, int *end){

         int left_sum, right_sum, sum, i, j;

         left_sum = array[mid];

         sum = 0;

         *begin = mid;

         *end = mid+1;

         for (i = mid; i >= low; i--){

                   sum += array[i];

                   if (sum > left_sum){

                            left_sum = sum;

                            *begin = i;

                   }

         }

         right_sum = array[mid+1];

         sum = 0;

         for (j = mid+1; j <= high; j++){

                   sum += array[j];

                   if (sum > right_sum){

                            right_sum = sum;

                            *end = j;

                   }

         }

         return left_sum + right_sum;

}

运行时间为θ(n)

只使用了一次for循环，注意，该算法只适用于数组原来的所有元素的求和值大于0，如果原来的数组的所有元素的求和值小于0,则不适用于本算法

C语言代码实现

#include<stdio.h>
#define LEN(array) (sizeof(array)/sizeof(array[0]))

void maxsubarray(int *array, int len, int *, int *);

main(){

//      int array[] = {13, -3, -25, 20, -3, -16, 23, -18, -20, -7, 12, -5, -22, 15, -4};

//      int array[] = {13, -3, -25};

         int array[] = {13, -3, -25, 55, -34};

         int i, start, end;
        

         printf("原始数组:\n");

         for (i = 0; i < LEN(array); i++)

                   printf("%5d", array[i]);
               

         putchar(‘\n‘);

         maxsubarray(array, LEN(array), &start, &end);

         printf("最大子序列的起始位置为：%d, 终止位置为：%d\n", start+1, end+1);

         printf("最大子序列为: ");

        
         for (i = start; i <= end; i++)

                   printf("%5d", array[i]);

         putchar(‘\n‘);

         putchar(‘\n‘);

}

 

void maxsubarray(int *array, int len, int *start, int *end){

         int i, sum, maxsum, temp;
      

         sum = maxsum = 0;

         temp = 0;

         for (i = 0; i < len; i++){

                   sum += array[i];

                   if (sum > maxsum){

                            maxsum = sum;

                            *end = i;

                            *start = temp;
                   }

                   else if (sum < 0){

                            //temp的值就是最大子序列的起始位置

                            sum = 0;

                            temp = i + 1;
                   }

         }       

         printf("\n子序列的求和最大值为： %d\n", maxsum);             

}