算法导论 4.1 最大子数组问题

 1 #include <iostream>
 2 using namespace std;
 3 
 4 const long long NINF = -1 << 30;
 5 const int N = 100;
 6 struct ans {
 7     long long sum;
 8     int left;
 9     int right;
10 };
11 
12 
13 ans FIND_CROSS_MAX_SUBARRAY(int a[], int low,int mid, int high) {
14     long long left_sum = NINF;
15     long long sum = 0;
16     int max_left;
17     
18     for (int i = mid; i >= low; i--) {
19         sum = sum + a[i];
20         if (sum > left_sum) {
21             left_sum = sum;
22             max_left = i;
23         }
24     }
25 
26     long long right_sum = NINF;
27     sum = 0;
28     int max_right;
29 
30     for (int j = mid + 1; j <= high; j++) {
31         sum += a[j];
32         if (sum > right_sum) {
33             right_sum = sum;
34             max_right = j;
35         }
36     }
37 
38     ans res{
39         right_sum + left_sum,max_left,max_right
40     };
41     return res;
42 }
43 
44 
45 ans FIND_MAX_SUBARRAY(int a[], int low, int high) {
46     if (low == high) {
47         ans res{ a[low],low,high };
48         return res;
49     }
50 
51     int mid = (low + high) / 2;
52     ans res1 = FIND_MAX_SUBARRAY(a, low, mid);
53     ans res2 = FIND_MAX_SUBARRAY(a, mid + 1, high);
54     ans res3 = FIND_CROSS_MAX_SUBARRAY(a, low,mid, high);
55 
56     if (res1.sum >= res2.sum&&res1.sum >= res3.sum)
57         return res1;
58     if (res2.sum >= res1.sum&&res2.sum >= res3.sum)
59         return res2;
60     if (res3.sum >= res1.sum&&res3.sum >= res2.sum)
61         return res3;
62 }
63 
64 
65 int main() {
66     int n;
67     int a[N];
68 
69     while (cin >> n) {
70         for (int i = 1; i <= n; i++)
71             cin >> a[i];
72         ans res = FIND_MAX_SUBARRAY(a, 1, n);
73         cout << "sum= " << res.sum << " [" << res.left << " , " << res.right << "]" << endl;
74     }
75     return 0;
76 }

 1 4. 1-3
 2 ans BRUTE_FIND_MAXIMUM_SUBARRAY(int a[], int low, int high) {
 3     int  sum = NINF;
 4     int left_i, right_j;
 5     ans res;
 6 
 7     for (int i = low; i <= high; i++) 
 8         for (int j = low; j <= high; j++) {
 9             int tmp = 0;
10             for (int k = i; k <= j; k++)
11                 tmp += a[k];
12             if (sum < tmp) {
13                 sum = tmp;
14                 left_i = i;
15                 right_j = j;
16             }
17         }
18     res.sum = sum;
19     res.right = right_j;
20     res.left = left_i;
21     return res;
22 }

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;
}