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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.
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)); } };
[LeetCode] Maximum Subarray 最大子数组
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原文地址：http://www.cnblogs.com/grandyang/p/4377150.html
