[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.

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.

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curSum，其中res保存最终要返回的结果，即最大的子数组之和，curSum初始值为0，每遍历一个数字num，比较curSum + num和num中的较大值存入curSum，然后再把res和curSum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：

C++ 解法一:

class Solution {
public:int maxSubArray(vector<int>& nums) {int res = INT_MIN, curSum = 0;for (int num : nums) {curSum = max(curSum + num, num);res = max(res, curSum);}return res;}
};

Java 解法一:

public class Solution {public int maxSubArray(int[] nums) {int res = Integer.MIN_VALUE, curSum = 0;for (int num : nums) {curSum = Math.max(curSum + num, num);res = Math.max(res, curSum);}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：

C++ 解法二:

class Solution {
public:int maxSubArray(vector<int>& nums) {if (nums.empty()) return 0;return helper(nums, 0, (int)nums.size() - 1);}int helper(vector<int>& nums, int left, int right) {if (left >= right) return nums[left];int mid = left + (right - left) / 2;int lmax = helper(nums, left, mid - 1);int rmax = helper(nums, mid + 1, right);int mmax = nums[mid], t = mmax;for (int i = mid - 1; i >= left; --i) {t += nums[i];mmax = max(mmax, t);}t = mmax;for (int i = mid + 1; i <= right; ++i) {t += nums[i];mmax = max(mmax, t);}return max(mmax, max(lmax, rmax));}
};

Java 解法二:

public class Solution {public int maxSubArray(int[] nums) {if (nums.length == 0) return 0;return helper(nums, 0, nums.length - 1);}public int helper(int[] nums, int left, int right) {if (left >= right) return nums[left];int mid = left + (right - left) / 2;int lmax = helper(nums, left, mid - 1);int rmax = helper(nums, mid + 1, right);int mmax = nums[mid], t = mmax;for (int i = mid - 1; i >= left; --i) {t += nums[i];mmax = Math.max(mmax, t);}t = mmax;for (int i = mid + 1; i <= right; ++i) {t += nums[i];mmax = Math.max(mmax, t);}return Math.max(mmax, Math.max(lmax, rmax));}
}

本文转自博客园Grandyang的博客，原文链接：最大子数组[LeetCode] Maximum Subarray ，如需转载请自行联系原博主。

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