最大子陣列和問題的解

QingLiXueShi發表於2015-07-07
陣列

尋找陣列A的和最大的非空連續子陣列。例如：A[] = {1, -2, 3, 10, -4, 7, 2, -5}的最大子陣列為3, 10, -4, 7, 2，其最大和為18。陣列元素都為負時，返回最大負值。

解法一：暴力列舉，O(N^2)

 1 //left,最大子陣列左端
 2 //right,最大子陣列右端
 3 //n陣列arr元素個數
 4 const int IntMin = (signed int)0x80000000;
 5 int MaxSum(int *arr, int &left, int &right, int n)
 6 {
 7     if (!arr || n <= 0) {
 8         left = -1;
 9         right = -1;
10         return IntMin;
11     }
12 
13     int maxSum = IntMin;
14     for (int i = 0; i < n; ++i) {
15         int sum = 0;
16         for (int j = i; j < n; ++j) {
17             sum += arr[j];
18             if (sum > maxSum) {
19                 left = i;
20                 right = j;
21                 maxSum = sum;
22             }
23         }
24     }
25     return maxSum;
26 }

解法二：分治，O(NlogN)

尋找陣列A[low…high]的最大子陣列時，將陣列劃分為兩個規模儘可能相等的子陣列 A[low…mid]和A[mid+1…high]。

A[low…high]的任何連續子陣列A[i…j]所處的位置必然是三種情況之一：

（1）完全位於子陣列A[low..mid]中，因此low<=i<=j<=mid; 

（2）完全位於子陣列A[mid + 1..high]中，因此mid<=i<=j<=high;

（3）跨越了中點，因此low<=i<=mid<j<=high;

A[low…high]的一個最大子陣列必然是完全位於A[low…mid]中、完全位於A[mid+1…high]中或者跨越中點的所有子陣列中和最大者。

 1 //left和right是返回值,未初始化,不能直接使用
 2 const int IntMin = (signed int)0x80000000;
 3 int MaxSumCross(int *arr, int &left, int mid, int &right, int n)
 4 {
 5     if (arr == NULL) {
 6         left = -1;
 7         right = -1;
 8         return IntMin;
 9     }
10 
11     int lmaxSum = IntMin;
12     int sum = 0;
13     for (int i = mid; i >= 0; --i) {
14         sum += arr[i];
15         if (sum > lmaxSum) {
16             left = i;
17             lmaxSum = sum;
18         }
19     }
20 
21     int rmaxSum = IntMin;
22     sum = 0;
23     for (int i = mid + 1; i < n; ++i) {
24         sum += arr[i];
25         if (sum > rmaxSum) {
26             right = i;
27             rmaxSum = sum;
28         }
29     }
30     return lmaxSum + rmaxSum;
31 }
32 
33 int MaxSum(int *arr, int &left, int &right, int n)
34 {
35     if (!arr || left > right || n <= 0) {
36         left = -1;
37         right = -1;
38         return IntMin;
39     }
40     if (left == right) 
41         return arr[left];
42 
43     //lleft,lright左子陣列的左端,右端
44     //rleft,rright右子陣列的左端,右端
45     //mleft,mright中間子陣列的左端,右端
46     int mid = left + (right - left) / 2;
47     int lleft = left, lright = mid;
48     int rleft = mid + 1, rright = right;
49     int mleft, mright;
50     int leftSum = MaxSum(arr, lleft, lright, n);
51     int midSum = MaxSumCross(arr, mleft, mid, mright, n);
52     int rightSum = MaxSum(arr, rleft, rright, n);
53 
54     if (leftSum >= rightSum && leftSum >= midSum) {
55         left = lleft;
56         right = lright;
57         return leftSum;
58     } else if (rightSum >= leftSum && rightSum >= midSum) {
59         left = rleft;
60         right = rright;
61         return rightSum;
62     } else if (midSum >= leftSum && midSum >= rightSum) {
63         left = mleft;
64         right = mright;
65         return midSum;
66     }
67 }

解法三：動態規劃，O(N)

 1 int MaxSum(int *arr, int &left, int &right, int n)
 2 {
 3     if (!arr || n <= 0) {
 4         left = -1;
 5         right = -1;
 6         return IntMin;
 7     }
 8 
 9     int nAll = arr[n - 1];
10     int nStart = arr[n - 1];
11     left = 0;
12     right = n - 1;
13     for (int i = n - 2; i >= 0; --i) {
14         if (nStart < 0) {
15             right = i;
16             nStart = 0;
17         }
18         nStart += arr[i];
19         if (nStart > nAll) {
20             left = i;
21             nAll = nStart;
22         }
23     }
24     return nAll;
25 }

 

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