[LeetCode] Maximum Subarray 最大子数组

 

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.

 

这道题让我们求最大子数组之和，并且要我们用两种方法来解，分别是O(n)的解法，还有用分治法Divide and Conquer Approach，这个解法的时间复杂度是O(nlgn)，那我们就先来看O(n)的解法，定义两个变量res和tmp，其中res保存最终要返回的结果，即最大的子数组之和，tmp是个临时变量，初始值为数组的第一个数，每遍历一个数字A[i]，比较tmp + A[i]和A[i]中的较大值存入tmp，然后再把res和tmp中的较大值存入res，以此类推直到遍历完整个数组，可得到最大子数组的值存在res中，代码如下：

解法一

// O(n) solution
class Solution {
public:
    int maxSubArray(int A[], int n) {
        int res = A[0], tmp = A[0];
        for (int i = 1; i < n; ++i) {
            tmp = max(tmp + A[i], A[i]);
            res = max(res, tmp);
        }
        return res;
    }
};

 

题目还要求我们用分治法Divide and Conquer Approach来解，这个分治法的思想就类似于二分搜索法，我们需要把数组一分为二，分别找出左边和右边的最大子数组之和，然后还要从中间开始向左右分别扫描，求出的最大值分别和左右两边得出的最大值相比较取最大的那一个，代码如下：

解法二

// Divide and conquer approach
class Solution {
public:
    int maxSubArray(int A[], int n) {
        return getMaxSubArray(A, 0, n - 1);
    }
    int getMaxSubArray(int A[], int left, int right) {
        if (left >= right) return A[left];
        int mid = (left + right) / 2;
        int lmax = getMaxSubArray(A, left, mid - 1);
        int rmax = getMaxSubArray(A, mid + 1, right);
        int mmax = A[mid], tmp = A[mid];
        for (int i = mid - 1; i >= left; --i) {
            tmp += A[i];
            mmax = max(mmax, tmp);
        }
        tmp = mmax;
        for (int i = mid + 1; i <= right; ++i) {
            tmp += A[i];
            mmax = max(mmax, tmp);
        }
        return max(mmax, max(lmax, rmax));
    }
};