leetcode 52:Maximum Subarray

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题目：

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.

思路：

首先想到的是用动态规划来做，因为动态规划跟前面计算的结果有关。

所以，对于给定的数组nums，我们可以定义一个相同大小的数组DP来存储当前元素对应的最大的连续元素之和。

以[−2,1,−3,4,−1,2,1,−5,4]举例分析。

当i=0时，DP[i]=-2;

当i=1时，DP[i]=1;

当i=2时，DP[i]=-2;

当i=3时，DP[i]=4;

......

很容易得出DP的递推式如下：

DP[i]=max(DP[i-1]+nums[i],nums[i]);

DP中最大值即为最后的返回值。

时间复杂度：O(n);

实现如下：

class Solution {public:int maxSubArray(vector<int>& nums) {int size = nums.size();int maxSum = nums[0];vector<int> DP(size, 0);DP[0] = nums[0];for (int i = 1; i < size; i++){DP[i] = max(DP[i - 1] + nums[i], nums[i]);if (DP[i]>maxSum) maxSum = DP[i];}return maxSum;}};

下面针对more practice，用分而治之的方法求解（参考：算法导论）：

分治策略是将数组分成两个规模尽可能相等的子数组，然后求解两个子数组。

但是最大子数组有可能出现在以下几种情况中：

完全位于子数组nums[low...mid]中；

完全位于子数组nums[mid...high]中；

跨越了中点；

因此我们需在上面三种情况中找最大者。

时间复杂度：O（nlgn）

实现如下：

class Solution {public:int maxSubArray(vector<int>& nums) {int size = nums.size();if (size == 1) return nums[0];return FindMaximumSubArray(nums,0,size-1);}private:int FindMaxCrossingSubArray(vector<int>& nums, int low, int mid, int high){int leftSum = -0XFFFF, rightSum = -0XFFFF, tempSum = 0;for (int i = mid; i >= low; i--){tempSum += nums[i];if (tempSum > leftSum) leftSum = tempSum;}tempSum = 0;for (int i = mid + 1; i <= high; i++){tempSum += nums[i];if (tempSum > rightSum) rightSum = tempSum;}return leftSum + rightSum;}int FindMaximumSubArray(vector<int>& nums, int low, int high){if (high == low) return nums[low];int mid = (low + high) / 2;int leftSum = FindMaximumSubArray(nums,low,mid);int rightSum = FindMaximumSubArray(nums, mid+1, high);int crossSum = FindMaxCrossingSubArray(nums, low, mid, high);return max(max(leftSum, rightSum), crossSum);}};

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