題目描述

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.

在一個數組中找到連續的子陣列（至少包含一個數字），這個陣列的總和最大。 例如，給定陣列[-2,1，-3,4，-1,2,1，-5,4]，連續的子陣列[4，-1,2,1]具有最大的和= 6。 更多的練習：如果你已經找到了O（n）的解決方案，嘗試使用分而治之的方法編碼另一個解決方案，這是更微妙的。

解題思路

從頭開始累加，直到和為負。此時前面這段不能給後面的串帶來正收益，應捨棄，sum清零，然後再開始統計最大的sum。演算法時間複雜度O(n)。

題目中說可以嘗試一下分治法，我是沒想出來，後來看了leetcode上大神的程式碼才恍然大悟知道要怎麼寫。

For each subarray, calculate four attributes:

mx (largest sum of this subarray), 
lmx(largest sum starting from the left most element), 
rmx(largest sum ending with the right most element), 
sum(the sum of the total subarray). 

演算法複雜度作者竟然說是O(n)，但是我覺得是O(logn)？

class Solution {
public:
    void maxSubArray(vector<int>& nums, int
 
 l, int r, int& mx, int& lmx, int& rmx, int& sum) {
        if (l == r) {
            mx = lmx = rmx = sum = nums[l];
        }
        else {
            int m = (l + r) / 2;
            int mx1, lmx1, rmx1, sum1;
            int mx2, lmx2, rmx2, sum2;
            maxSubArray(nums, l, m, mx1, lmx1, rmx1, sum1);
            maxSubArray(nums, m + 1, r, mx2, lmx2, rmx2, sum2);
            mx = max(max(mx1, mx2), rmx1 + lmx2);
            lmx = max(lmx1, sum1 + lmx2);
            rmx = max(rmx2, sum2 + rmx1);
            sum = sum1 + sum2;
        }
    }
    int maxSubArray(vector<int>& nums) {
        if (nums.size() == 0) {
            return 0;
        }
        int mx, lmx, rmx, sum;
        maxSubArray(nums, 0, nums.size() - 1, mx, lmx, rmx, sum);
        return mx;
    }
};

C++版程式碼實現

class Solution {
public:
    int maxSubArray(int A[], int n) {
        if(A == NULL || n == 0)
            return false;
        int sum = A[0];
        int maxSum = A[0];
        for(int i = 1; i < n; ++i){
            if(sum < 0)
                sum = 0;
            sum += A[i];
            maxSum = max(sum, maxSum);
        }
        return maxSum;
    }
};

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