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.

這道題讓我們求最大子陣列之和，並且要我們用兩種方法來解，分別是O(n)的解法，還有用分治法Divide and Conquer Approach，這個解法的時間複雜度是O(nlgn)，那我們就先來看O(n)的解法，定義兩個變數res和curSum，其中res儲存最終要返回的結果，即最大的子陣列之和，curSum初始值為0，每遍歷一個數字num，比較curSum + num和num中的較大值存入curSum，然後再把res和curSum中的較大值存入res，以此類推直到遍歷完整個陣列，可得到最大子陣列的值存在res中，程式碼如下：

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;
    }
}

題目還要求我們用分治法Divide and Conquer Approach來解，這個分治法的思想就類似於二分搜尋法，我們需要把陣列一分為二，分別找出左邊和右邊的最大子陣列之和，然後還要從中間開始向左右分別掃描，求出的最大值分別和左右兩邊得出的最大值相比較取最大的那一個，程式碼如下：

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));
    }
}