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.

Solution1：DP

思路：
dp[i] means the maximum subarray ending with A[i];
dp[i] = A[i] + (dp[i - 1] > 0 ? dp[i - 1] : 0); 当前面以i-1结尾的最大和dp[i-1]为负时，也就是加了前面的 还没有自己大的情况下，dp[i] = 自己A[i]，反之继续累加。
Time Complexity: O(N) Space Complexity: O(N)

实现1_b: P with Space Optimization，只依赖与dp[i - 1] 所以可以优化空间
Time Complexity: O(N) Space Complexity: O(1)

Solution2：在累积sum(离散积分) 上找 递增上"/"的最大差

思路： 累积sum 上的 递增的最大差 是就 原序列连续子序列的最大和。

Solution3：Divide & Conquer 分治

思路： (Not finished)

Solution1a Code：

class Solution1a {
    public int maxSubArray(int[] nums) {
            int n = nums.length;
            int[] dp = new int[n]; //dp[i] means the maximum subarray ending with A[i];
            dp[0] = nums[0];
            int g_max = nums[0];

            for(int i = 1; i < n; i++){
                dp[i] = nums[i] + (dp[i - 1] > 0 ? dp[i - 1] : 0);
                if(dp[i] > g_max) {
                    g_max = dp[i];
                }
            }

            return g_max;
    }
}

Solution1b Code：

class Solution1b {
    public int maxSubArray(int[] nums) {
        int cur_max = nums[0];
        int g_max = nums[0];
        
        for(int i = 1; i < nums.length; i++) {
            cur_max = (nums[i] > cur_max + nums[i]) ? nums[i] : cur_max + nums[i];
            if(cur_max > g_max) {
                g_max = cur_max;
            }
        }
        return g_max;
    }
}

Solution2 Code：

class Solution2 {
    public int maxSubArray(int[] nums) {
        int accusum = 0, prev_g_min = 0;
        int g_max = nums[0];
        
        for(int i = 0; i < nums.length; i++) {
            accusum += nums[i];
            if(accusum - prev_g_min > g_max) {
                g_max = accusum - prev_g_min;
            }
            // update prev_g_min
            if(accusum < prev_g_min) {
                prev_g_min = accusum;
            }
        }
        return g_max;
    }
}