最大子序列问题

最大子数组问题

定义

给定整数A1, A2, …, An(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);

}