LeetCode:Maximum Subarray
Find the contiguous subarray within an array (containing at least one number) which has the largest sum.
For example, given the array [−2,1,−3,4,−1,2,1,−5,4]
,
the contiguous subarray [4,−1,2,1]
has the largest sum = 6
.
More practice:
If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.
可以参考我的另一篇博文最大子数组和(最大子段和)。
下面分别给出O(n)的动态规划解法和O(nlogn)的分治解法 本文地址
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 | class Solution { public : int maxSubArray( int A[], int n) { //最大字段和问题 int res = INT_MIN, sum = -1; for ( int i = 0; i < n; i++) { if (sum > 0) sum += A[i]; else sum = A[i]; if (sum > res)res = sum; } return res; } }; |
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 | class Solution { public : int maxSubArray( int A[], int n) { //最大字段和问题 return helper(A, 0, n-1); } private : int helper( int A[], const int istart, const int iend) { if (istart == iend) return A[iend]; int middle = (istart + iend) / 2; int maxLeft = helper(A, istart, middle); int maxRight = helper(A, middle + 1, iend); int midLeft = A[middle]; int tmp = midLeft; for ( int i = middle - 1; i >= istart; i--) { tmp += A[i]; if (midLeft < tmp)midLeft = tmp; } int midRight = A[middle + 1]; tmp = midRight; for ( int i = middle + 2; i <= iend; i++) { tmp += A[i]; if (midRight < tmp)midRight = tmp; } return max(max(maxLeft, maxRight), midLeft + midRight); } }; |
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