[LC] 53. Maximum Subarray

Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.

Example:

Input: [-2,1,-3,4,-1,2,1,-5,4],
Output: 6
Explanation: [4,-1,2,1] has the largest sum = 6.

Time:O(N)
Space:O(N)

//prefix

public class Solution {
    /**
     * @param nums: A list of integers
     * @return: A integer indicate the sum of max subarray
     */
    public int maxSubArray(int[] nums) {
        // write your code here
        if (nums == null || nums.length == 0) {
            return 0;
        }
        int max = Integer.MIN_VALUE, min = 0, sum = 0;
        for (int i = 0; i < nums.length; i++) {
            sum += nums[i];
            max = Math.max(max, sum - min);
            min = Math.min(min, sum);
        }
        return max;
    }
}

//dp

 class Solution:
     def maxSubArray(self, nums: List[int]) -> int:
         if nums is None or len(nums) == 0:
             return

         sum_arr = [0] * len(nums)
         sum_arr[0] = nums[0]
         max_res = nums[0]
         for i in range(1, len(nums)):
             if sum_arr[i - 1] > 0:
                 sum_arr[i] = nums[i] + sum_arr[i - 1]
             else:
                 sum_arr[i] = nums[i]
             max_res = max(max_res, sum_arr[i])
         return max_res

class Solution {
    public int maxSubArray(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        int sum = nums[0], res = sum;
        for (int i = 1; i < nums.length; i++) {
            // add sum if sum > 0
            sum = Math.max(nums[i], sum + nums[i]);
            res = Math.max(res, sum);
        }
        return res;
    }
}