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.

click to show more practice.

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.

1.linear time method.

 

2.

public class Solution {
    public int maxSubArray(int[] A) {
        int len = A.length;
        int curr = A[0], next = A[0];
        for(int i = 1; i < len; ++i){
            next = (next <= 0) ? A[i] : next + A[i];
            curr = Math.max(curr, next); 
        }
        return curr;   
    }
}

　　

divide and conquer method.

public class Solution {
	/**
	 * This program is an implementation of divide-and-conquer method
	 * @param A --an array of integer
	 * @return the maximum sum of a subarray
	 * @author Averill Zheng
	 * @version 2014-06-04
	 * @since JDK 1.7
	 */
	public int maxSubArray(int[] A) {
        return maxSubArrayHelper(A, 0, A.length);
    }
	/**
	 * The function is used to find the maximum sum of a contiguous subarray between index i (inclusive) and j
	 * (exclusive).
	 * @param A --an integer array.
	 * @param i --start index which is included.	  
	 * @param j --end index which is exclusive.
	 * @return int --maximum sum of a contiguous subarray.
	 */
	private int maxSubArrayHelper(int[] A, int i, int j){
		int max = Integer.MIN_VALUE;
		if(j > i){
			if(j == i + 1)
				max = A[i];
			else{
				int mid = i + (j - i) / 2;
				int leftMax = maxSubArrayHelper(A, i, mid);
				int rightMax = maxSubArrayHelper(A,mid, j);
				int crossMax = crossMax(A, i, j, mid);
				max = Math.max(Math.max(leftMax, crossMax), rightMax);
			}
		}
		return max;
	}
	private int crossMax(int[] A, int i, int j, int mid){
		//[i,mid) is the first half and [mid, j) is the second half
		//both of them contains at least one element
		
		//left max
		int leftMax = A[mid - 1];
		int leftSum = A[mid - 1];
		for(int k = mid - 2; k > i - 1; --k){
			leftSum += A[k];
			leftMax = (leftMax < leftSum) ? leftSum : leftMax;
		}
		
		//right max
		int rightMax = A[mid];
		int rightSum = A[mid];
		for(int k = mid + 1; k < j; ++k){
			rightSum += A[k];
			rightMax = (rightMax < rightSum) ? rightSum : rightMax; 
		}
		return leftMax + rightMax;
	}
}