Leetcode Maximum Subarray

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.

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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.

这是个很经典的题目了。

两个方法：

1 动态规划法

2 二分法

这里是使用简单的动态规划法，只需要一个额外空间记录就可以了，只要知道方法，实现起来比二分法简单多了。

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

二分法关键： 当最大子序列出现在中间的时候，应该如何计算最大值？

1 要循环完左边和右边的序列

2 从中间到左边，找到最大值

3 从中间到右边，找到最大值

这个判断条件还是有点难想的。

不过本题卡我时间最长的居然是一个小问题：最左边的下标写成了0.狂晕！

class Solution {
public:
	int maxSubArray(int A[], int n) {
		if (n < 1) return 0;
		if (n == 1) return A[0];
		return biSubArray(A, 0, n-1);
	}

	//low和up均为C++存储内容的下标
	int biSubArray(int A[], int low, int up)
	{
		if (low > up) return INT_MIN;
		if (low == up) return A[low];

		int mid = low+((up-low)>>1);

		//low不小心写成了0，结果浪费了很长时间debug
		int lSum = biSubArray(A, low, mid);
		int rSum = biSubArray(A, mid+1, up);

		int midLSum = INT_MIN;
		int tempSum = 0;

		//low不小心写成了0，结果浪费了很多时间调试。
		for (int i = mid; i >= low; i--)
		{
			tempSum += A[i];
			//注意：条件的构造
			if (tempSum > midLSum)
				midLSum = tempSum;
		}

		int midRSum = INT_MIN;
		tempSum = 0;
		for (int i = mid+1; i <= up; i++)
		{
			tempSum += A[i];
			if (tempSum > midRSum)
				midRSum = tempSum;
		}

		int midSum = midLSum + midRSum;

		return max(max(lSum,rSum), midSum);
	}
};