Maximum subarray@leetcode

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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.

» Solve this problem

The first way to do it is to iterate through the whole array and use two variables, sum and max to dynamically record the max subarray value.

Cases can be concluded below,
1. All elements in array are negative, we just need to find the biggest one.
2. Have both negative and positive values.
    (1) discard negative ones, while we need to find the maximum subarrays, so it possible that the maximum subarray includes both negative and positive values. In this case, we use a sum to record.
     We use max to represent that maximum value of subarray. Assume that currently we have max>0, now we go ahead and scan. If A[i]<0, discard, go ahead. But if A[i]+A[i+1]+...+A[i+k]>0, then we need add Ai..k to subarray, and replace max = max + A[i]+...+A[i+k]. Based on the analysis above, we can have the algorithm given below.

	class Solution {
	public:
	int maxSubArray(int A[], int n) {
	int sum = 0, max = INT_MIN;
	for(int i=0; i<n; i++){
	sum+=A[i];
	if(sum>max)
	max = sum;
	if(sum<0)
	sum = 0;
	}
	return max;
	}
	};

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The second way to do it is by divide and conquer. 

	int maxSubArray(int A[], int n) {
	2: // Start typing your C/C++ solution below
	3: // DO NOT write int main() function
	4: int maxV = INT_MIN;
	5: return maxArray(A, 0, n-1, maxV);
	6: }
	7: int maxArray(int A[], int left, int right, int& maxV)
	8: {
	9: if(left>right)
	10: return INT_MIN;
	11: int mid = (left+right)/2;
	12: int lmax = maxArray(A, left, mid -1, maxV);
	13: int rmax = maxArray(A, mid + 1, right, maxV);
	14: maxV = max(maxV, lmax);
	15: maxV = max(maxV, rmax);
	16: int sum = 0, mlmax = 0;
	17: for(int i= mid -1; i>=left; i--)
	18: {
	19: sum += A[i];
	20: if(sum > mlmax)
	21: mlmax = sum;
	22: }
	23: sum = 0; int mrmax = 0;
	24: for(int i = mid +1; i<=right; i++)
	25: {
	26: sum += A[i];
	27: if(sum > mrmax)
	28: mrmax = sum;
	29: }
	30: maxV = max(maxV, mlmax + mrmax + A[mid]);
	31: return maxV;
	32: }

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	int maxSubArray(int A[], int n) {
	vector<int> rec(n, 0);
	rec[0] = A[0];
	int max = rec[0];
	for(int i=1; i<n; i++){
	if(rec[i-1]+A[i] < A[i]) rec[i] = A[i];
	else
	rec[i] = rec[i-1] + A[i];
	if(rec[i]>max) max = rec[i];
	}
	return max;
	}

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