分治法求最大子数组
1 /**
2 *分治法求最大子数组
3 *yaoC 2014-09-06
4 *Θ(nlg(n))
5 */
6
7 #include<iostream>
8 using namespace std;
9
10 #define neg_infinity -999999
11
12 //一个子数组的开始下标，结束下标和值的和
13 struct indices_and_value
14 {
15 int left;
16 int right;
17 int value;
18 };
19
20 /**
21 * [findMaxCrossingSubarray 找到跨越中点的最大子数组]
22 * @param A [输入数组]
23 * @param low [数组开始下标]
24 * @param mid [数组中间下标]
25 * @param heigh [数组结束下标]
26 * @return [所求子数组的开始结束下标、值的和]
27 */
28 indices_and_value findMaxCrossingSubarray(int* A, int low, int mid, int heigh)
29 {
30 int sum = 0;
31 int left_sum = neg_infinity;//最大子数组在中点左边的值的和
32 indices_and_value p;
33 for (int i = mid; i >= low; i--)
34 {
35 sum += A[i];
36 if (sum>left_sum)
37 {
38 left_sum = sum;
39 p.left = i;
40 }
41 }
42 sum = 0;
43 int right_sum = neg_infinity;//最大子数组在中点右边的值的和
44 for (int i = mid + 1; i <= heigh; i++)
45 {
46 sum += A[i];
47 if (sum>right_sum)
48 {
49 right_sum = sum;
50 p.right = i;
51 }
52 }
53 p.value = left_sum + right_sum;
54 return p;
55 }
56
57 /**
58 * [findMaxSubarray 找到非跨越中点的最大子数组]
59 * @param A [输入数组]
60 * @param low [数组开始下标]
61 * @param heigh [数组结束下标]
62 * @return [所求子数组的开始结束下标、值的和]
63 */
64 indices_and_value findMaxSubarray(int* A, int low, int heigh)
65 {
66 int mid = (low + heigh) / 2;//数组中点
67 indices_and_value p;
68 if (low == heigh)//base case:only one element
69 {
70 p.left = low;
71 p.right = low;
72 p.value = A[low];
73 }
74 else//recursive case
75 {
76 indices_and_value left = findMaxSubarray(A, low, mid);
77 indices_and_value cross = findMaxCrossingSubarray(A,low,mid,heigh);
78 indices_and_value right = findMaxSubarray(A,mid + 1,heigh);
79 if (left.value>cross.value)
80 p = left;
81 else
82 {
83 if (right.value>cross.value)
84 p = right;
85 else p = cross;
86 }
87 }
88 return p;
89 }
90
91 int main()
92 {
93 int test[] = { 0, 13, -3, -25, 20, -3, -16, -23, 18, 20, -7, 12, -5, -22, 15, -4, 7 };
94 indices_and_value p;
95 p = findMaxSubarray(test, 0, 16);//(8,11,43)
96 cout <<"("<< p.left << "," << p.right << "," << p.value <<")"<< endl;
97 }
2 *分治法求最大子数组
3 *yaoC 2014-09-06
4 *Θ(nlg(n))
5 */
6
7 #include<iostream>
8 using namespace std;
9
10 #define neg_infinity -999999
11
12 //一个子数组的开始下标，结束下标和值的和
13 struct indices_and_value
14 {
15 int left;
16 int right;
17 int value;
18 };
19
20 /**
21 * [findMaxCrossingSubarray 找到跨越中点的最大子数组]
22 * @param A [输入数组]
23 * @param low [数组开始下标]
24 * @param mid [数组中间下标]
25 * @param heigh [数组结束下标]
26 * @return [所求子数组的开始结束下标、值的和]
27 */
28 indices_and_value findMaxCrossingSubarray(int* A, int low, int mid, int heigh)
29 {
30 int sum = 0;
31 int left_sum = neg_infinity;//最大子数组在中点左边的值的和
32 indices_and_value p;
33 for (int i = mid; i >= low; i--)
34 {
35 sum += A[i];
36 if (sum>left_sum)
37 {
38 left_sum = sum;
39 p.left = i;
40 }
41 }
42 sum = 0;
43 int right_sum = neg_infinity;//最大子数组在中点右边的值的和
44 for (int i = mid + 1; i <= heigh; i++)
45 {
46 sum += A[i];
47 if (sum>right_sum)
48 {
49 right_sum = sum;
50 p.right = i;
51 }
52 }
53 p.value = left_sum + right_sum;
54 return p;
55 }
56
57 /**
58 * [findMaxSubarray 找到非跨越中点的最大子数组]
59 * @param A [输入数组]
60 * @param low [数组开始下标]
61 * @param heigh [数组结束下标]
62 * @return [所求子数组的开始结束下标、值的和]
63 */
64 indices_and_value findMaxSubarray(int* A, int low, int heigh)
65 {
66 int mid = (low + heigh) / 2;//数组中点
67 indices_and_value p;
68 if (low == heigh)//base case:only one element
69 {
70 p.left = low;
71 p.right = low;
72 p.value = A[low];
73 }
74 else//recursive case
75 {
76 indices_and_value left = findMaxSubarray(A, low, mid);
77 indices_and_value cross = findMaxCrossingSubarray(A,low,mid,heigh);
78 indices_and_value right = findMaxSubarray(A,mid + 1,heigh);
79 if (left.value>cross.value)
80 p = left;
81 else
82 {
83 if (right.value>cross.value)
84 p = right;
85 else p = cross;
86 }
87 }
88 return p;
89 }
90
91 int main()
92 {
93 int test[] = { 0, 13, -3, -25, 20, -3, -16, -23, 18, 20, -7, 12, -5, -22, 15, -4, 7 };
94 indices_and_value p;
95 p = findMaxSubarray(test, 0, 16);//(8,11,43)
96 cout <<"("<< p.left << "," << p.right << "," << p.value <<")"<< endl;
97 }
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