資料結構與演算法分析（c 語言描述）最大子序列和問題

ps：我怕我發的會打擾到大家，本意只是想記下學習研究過程跟結果，這些很基礎的東西，如果打擾到大家的話在底下說下，以後就不發這種了

/**
 * 給定(可能有負數)整數a(1)、a(2)、……a(n)，求 a(1)+a(2)+……+a(j)的最大值。
 * 為方便起見，若所有的整數為負數，則最大子序列和為0.
 * 描述：在一系列整數中，找出連續的若干個整數，這若干個整數之和 最大。
 * 
 * 四個演算法，充分體現出優秀的演算法的重要性
 */

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

// #define N 100

int bigO_n3(int arr[], int n);

int bigO_n2(int arr[], int n);

int bigO_logn(int arr[], int left, int right);

int bigO_n(int arr[], int n);
// 第三個演算法用到
int max(int a, int b, int c);

int main(void)
{
    int * arr;
    clock_t elapse;

    printf("\t\tO(n^3)\t\tO(n^2)\t\tO(logn)\t\tO(n)\n");
    printf("\tN\t耗時\t\t耗時\t\t耗時\t\t耗時\n");

    for (int N = 10; N <= 10000; N = N*2) {
        srand((unsigned)time(NULL));
        arr = (int *) malloc(sizeof(int) * N);
        for (int j = 0; j < N; j++) {
            arr[j] = rand() % 2001 - 1000;
        }

        printf("\t%d", N);

        elapse = clock();
        bigO_n3(arr, N);
        elapse = clock() - elapse;
        printf("\t%.6lf", (double) elapse / CLOCKS_PER_SEC );

        elapse = clock();
        bigO_n2(arr, N);
        elapse = clock() - elapse;
        printf("\t%.6lf", (double) elapse / CLOCKS_PER_SEC );

        elapse = clock();
        bigO_logn(arr, 0, N - 1);
        elapse = clock() - elapse;
        printf("\t%.6lf", (double) elapse / CLOCKS_PER_SEC );

        elapse = clock();
        bigO_n(arr, N);
        elapse = clock() - elapse;
        printf("\t%.6lf", (double) elapse / CLOCKS_PER_SEC );

        printf("\n");

        free(arr);
    }

    // bigO_n3()
    return 0;
}

int bigO_n3(int arr[], int n)
{
    int currenSum, maxSum, i, j, k;
    maxSum = 0;

    for (i = 0; i < n; i++) {
        for(j = i; j < n; j++) {
            currenSum = 0;
            for (k = i; k <= j; k++)
                currenSum += arr[k];

            if (currenSum > maxSum)
                maxSum = currenSum;
        }
    }
    return maxSum;
}

int bigO_n2(int arr[], int n)
{
    int currenSum, maxSum, i, j;
    maxSum = 0;

    for (i = 0; i< n; i++) {
        currenSum = 0;
        for( j = i; j < n; j++) {
            currenSum += arr[j];

            if (currenSum > maxSum)
                maxSum = currenSum;
        }
    }
    return maxSum;
}

int bigO_logn(int arr[], int left, int right)
{
    int currenSum, maxSum, j, center;
    int leftBorderSum, rightBorderSum;
    int maxLeftBorderSum, maxRightBorderSum;
    int maxLeftSum, maxRightSum;

    // 單個數
    if ( left == right) {
        if (arr[left] > 0)
            return arr[left];
        return 0;
    }

    //分
    center = (int)(left + right) / 2;
    //最大值
    maxLeftSum = bigO_logn(arr, left, center);
    maxRightSum = bigO_logn(arr, center + 1, right);

    maxLeftBorderSum = 0; leftBorderSum = 0;
    for (j = center; j >= 0; j--) {
        leftBorderSum += arr[j];
        if (leftBorderSum >  maxLeftBorderSum)
            maxLeftBorderSum = leftBorderSum;
    }

    maxRightBorderSum = 0; rightBorderSum = 0;
    for (j = center + 1; j <= right; j++) {
        rightBorderSum += arr[j];
        if (rightBorderSum > maxRightBorderSum)
            maxRightBorderSum = rightBorderSum;
    }

    return max(maxLeftSum, maxRightSum, maxLeftBorderSum + maxRightBorderSum);
}

int bigO_n(int arr[], int n)
{
    int i, currenSum, maxSum;
    currenSum = maxSum = 0;
    for (i = 0; i < n; i++) {
        currenSum += arr[i];

        if (currenSum > maxSum)
            maxSum = currenSum;

        if (currenSum < 0)
            currenSum = 0;
    }
    return maxSum;
}

int max(int a, int b, int c)
{
    if (a >= b && a >= c) {
        return a;
    } else if (b >= a && b >= c) {
        return b;
    } else {
        return c;
    }
}

執行結果