LeetCode

Description

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.
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.

Solution

解题思路参考了

1. Kadane算法

2. @LiamHuang said in DP solution & some thoughts:

   When we are talking about DP, the first problem comes out to our mind should be: what's the statement of the sub-problem, whose format should satisfy that if we've solved a sub-problem, it would be helpful for solving the next-step sub-problem, and, thus, eventually helpful for solving the original problem.

   Here is the sub-problem we state: denote int_local_max[i] as the max-sub-array-sum that ends with nums[i]. The relationship between the two steps is simple: int_local_max[i + 1] = max (int_local_max[i] + nums[i + 1], nums[i + 1]) or int_local_max[i + 1] = (int_local_max[i] > 0) ? int_local_max[i] + nums[i + 1] : nums[i + 1].

   Now, all we have to do is to scan through the array, and find which int_local_max[i] is the maximum of all the int_local_maxs.

python

class Solution(object):
    def maxSubArray(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        # [-2,1,-3,4,-1,2,1,-5,4]
        max_num = max(nums)
        if max_num < 0:
            return max_num

        max_sum = 0
        cur_sum = 0
        nums_len = len(nums)
        for i in range(nums_len):
            cur_sum = cur_sum + nums[i]
            if cur_sum < 0:
                cur_sum = 0
                continue

            if cur_sum > max_sum:
                max_sum = cur_sum

        return max_sum

cpp

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        if (nums.size() == 0)
            return 0;

        if (nums.size() == 1)
            return nums.at(0);

        int max_sum = nums.at(0);
        int cur_sum = nums.at(0);
        size_t length = nums.size();
        for (size_t i=1; i<length; ++i)
        {
            cur_sum = max(cur_sum + nums.at(i), nums.at(i));
            max_sum = max(cur_sum, max_sum);
        }

        return max_sum;
    }
}; 

Reference

• http://blog.csdn.net/joylnwang/article/details/6859677
• https://discuss.leetcode.com/topic/6413/dp-solution-some-thoughts