【推荐】2019 Java 开发者跳槽指南.pdf(吐血整理) >>>

复杂度1 最大子列和问题 

给定KKK个整数组成的序列{ N1N_1N1, N2N_2N2, ..., NKN_KNK }，“连续子列”被定义为{ NiN_iNi, Ni+1N_{i+1}Ni+1, ..., NjN_jNj }，其中 1≤i≤j≤K1 \le i \le j \le K1≤i≤j≤K。“最大子列和”则被定义为所有连续子列元素的和中最大者。例如给定序列{ -2, 11, -4, 13, -5, -2 }，其连续子列{ 11, -4, 13 }有最大的和20。现要求你编写程序，计算给定整数序列的最大子列和。

本题旨在测试各种不同的算法在各种数据情况下的表现。各组测试数据特点如下：

• 数据1：与样例等价，测试基本正确性；

• 数据2：10²个随机整数；

• 数据3：10³个随机整数；

• 数据4：10⁴个随机整数；

• 数据5：10⁵个随机整数；

输入格式:

输入第1行给出正整数KKK (≤100000\le 100000≤100000)；第2行给出KKK个整数，其间以空格分隔。

输出格式:

在一行中输出最大子列和。如果序列中所有整数皆为负数，则输出0。

输入样例:

6
-2 11 -4 13 -5 -2

输出样例:

20

CODE：

#include <stdio.h>
#define LEN 100000
 
int maxSubSeqSum( int a[], int n );
void printR( int result );

int main()
{
    int n;
    scanf("%d",&n);
    int a[LEN];
    int i;
    for( i=0; i<n; i++ ){
    	scanf("%d",&a[i]);
    }
    printR(maxSubSeqSum( a, n ));
    return 0;
}

int maxSubSeqSum( int a[], int n )
{
    int ret=0;
    int thisSum=0;
    int i;
    for( i=0; i<n; i++ ){
        thisSum+=a[i];
        if( thisSum>=ret ){
            ret=thisSum;
        }else if( thisSum<0 ){
            thisSum=0;
        }
    }
    return ret;
}

void printR( int result )
{
    if( result ){
        printf("%d\n",result);
    }else{
        printf("0\n");	
    }
}

复杂度2 Maximum Subsequence Sum

Given a sequence of KKK integers { N1N_1N1, N2N_2N2, ..., NKN_KNK }. A continuous subsequence is defined to be { NiN_iNi, Ni+1N_{i+1}Ni+1, ..., NjN_jNj } where 1≤i≤j≤K1 \le i \le j \le K1≤i≤j≤K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For example, given sequence { -2, 11, -4, 13, -5, -2 }, its maximum subsequence is { 11, -4, 13 } with the largest sum being 20.

Now you are supposed to find the largest sum, together with the first and the last numbers of the maximum subsequence.

Input Specification:

Each input file contains one test case. Each case occupies two lines. The first line contains a positive integer KKK (≤10000\le 10000≤10000). The second line contains KKK numbers, separated by a space.

Output Specification:

For each test case, output in one line the largest sum, together with the first and the last numbers of the maximum subsequence. The numbers must be separated by one space, but there must be no extra space at the end of a line. In case that the maximum subsequence is not unique, output the one with the smallest indices iii and jjj (as shown by the sample case). If all the KKK numbers are negative, then its maximum sum is defined to be 0, and you are supposed to output the first and the last numbers of the whole sequence.

Sample Input:

10
-10 1 2 3 4 -5 -23 3 7 -21

Sample Output:

10 1 4  

CODE：

#include <stdio.h>

#define LEN 100000

int left,right;
int maxSubSeqSum( int a[], int n );
void printR( int result );

int main()
{
    int n;
    scanf("%d",&n);
    int a[LEN];
    int i;
    for( i=0; i<n; i++ ){
        scanf("%d",&a[i]);
    }
    left=a[0];
    right=a[n-1];
    printR(maxSubSeqSum( a, n ));
    return 0;
}


int maxSubSeqSum( int *a, int n )
{
    int ret=0;
    int thisSum=0;
    int i;
    int ltemp,rtemp;
    int flag=1;
    for( i=0; i<n; i++ ){
        thisSum+=a[i];
        if( thisSum>ret ){
            ret=thisSum;
            right=a[i];
            ltemp=left;
            rtemp=right;
        }else if( thisSum==0&&ret==0 ){
            right=a[i];
            ltemp=left;
            rtemp=right;
            flag=0;
        }else if( thisSum<0 ){
            thisSum=0;
            left=a[i+1];
        }
    }
    if( ret==0&&flag ){
        left=a[0];
        right=a[n-1];
    }else if(ret!=0||!flag){
        left=ltemp;
        right=rtemp;
    }
    return ret;
}

void printR( int result )
{
    if( result ){
        printf("%d ",result);
    }else{
        printf("0 ");	
    }
    printf("%d ",left);
    printf("%d\n",right);
}