思路1

1.暴力解法，两重循环，从某一个下标开始，计算连续的和，求最大

代码

class Solution
{
public:
    int maxSubArray(vector<int> &nums)
    {
        //类似寻找最大最小值的题目，初始值一定要定义成理论上的最小最大值
        int max = INT_MIN;
        int numsSize = int(nums.size());
        for (int i = 0; i < numsSize; i++)
        {
            int sum = 0;
            for (int j = i; j < numsSize; j++)
            {
                sum += nums[j];
                if (sum > max)
                {
                    max = sum;
                }
            }
        }
        return max;
    }
};

思路2 动态规划

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int pre = 0,Max_val = nums[0];
        for(const auto &x:nums) //需要改变迭代对象  加&
        {
            pre = max(pre+x,x); //选择每一步加入的值进行比较，去较大的值作为加入当前的子序列
            Max_val = max(Max_val,pre); 
        }
        return Max_val;
    }
};

思路3

class Solution(object):
    def maxSubArray(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        len_num = len(nums)
        for i in range(1,len_num):
            nums[i] = max(nums[i],nums[i]+nums[i-1])
        return max(nums)