算法导论 4.1 最大子数组问题

8. 实现代码

#include <iostream>
using namespace std;

const long long NINF = -1 << 30;
const int N = 100;
struct ans {
    long long sum;
    int left;
    int right;
};


ans FIND_CROSS_MAX_SUBARRAY(int a[], int low,int mid, int high) {
    long long left_sum = NINF;
    long long sum = 0;
    int max_left;
    
    for (int i = mid; i >= low; i--) {
        sum = sum + a[i];
        if (sum > left_sum) {
            left_sum = sum;
            max_left = i;
        }
    }

    long long right_sum = NINF;
    sum = 0;
    int max_right;

    for (int j = mid + 1; j <= high; j++) {
        sum += a[j];
        if (sum > right_sum) {
            right_sum = sum;
            max_right = j;
        }
    }

    ans res{
        right_sum + left_sum,max_left,max_right
    };
    return res;
}


ans FIND_MAX_SUBARRAY(int a[], int low, int high) {
    if (low == high) {
        ans res{ a[low],low,high };
        return res;
    }

    int mid = (low + high) / 2;
    ans res1 = FIND_MAX_SUBARRAY(a, low, mid);
    ans res2 = FIND_MAX_SUBARRAY(a, mid + 1, high);
    ans res3 = FIND_CROSS_MAX_SUBARRAY(a, low,mid, high);

    if (res1.sum >= res2.sum&&res1.sum >= res3.sum)
        return res1;
    if (res2.sum >= res1.sum&&res2.sum >= res3.sum)
        return res2;
    if (res3.sum >= res1.sum&&res3.sum >= res2.sum)
        return res3;
}


int main() {
    int n;
    int a[N];

    while (cin >> n) {
        for (int i = 1; i <= n; i++)
            cin >> a[i];
        ans res = FIND_MAX_SUBARRAY(a, 1, n);
        cout << "sum= " << res.sum << " [" << res.left << " , " << res.right << "]" << endl;
    }
    return 0;
}

4. 1-3
ans BRUTE_FIND_MAXIMUM_SUBARRAY(int a[], int low, int high) {
    int  sum = NINF;
    int left_i, right_j;
    ans res;

    for (int i = low; i <= high; i++) 
        for (int j = low; j <= high; j++) {
            int tmp = 0;
            for (int k = i; k <= j; k++)
                tmp += a[k];
            if (sum < tmp) {
                sum = tmp;
                left_i = i;
                right_j = j;
            }
        }
    res.sum = sum;
    res.right = right_j;
    res.left = left_i;
    return res;
}

实现代码
ans LINER_TIME_FIND_MAXIMUM_SUBARRAY(int a[], int n) {
    int sum = a[1], Start = 1, End = 1;
    int tsum = a[1], tstart = 1, tend = 1;

    for (int i = 2; i <= n; i++) {
        if (tsum < 0) {
            tstart = tend = i;
            tsum = a[i];
        }
        else {
            tend = i;
            tsum += a[i];
        }
        if (tsum > sum) {
            sum = tsum;
            Start = tstart;
            End = tend;
        }
    }
    ans res{
        sum,Start,End
    };
    return res;
}