LeetCode Array Easy 53. Maximum Subarray 个人解法和分治思想的学习

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.

Follow up:

If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.

我的解法：两层循环第一层循环通过i控制进度，第二层循环计算从i 开始的子数组的和的最大值

C#版解法：

public class Solution {
    public int MaxSubArray(int[] nums) {
        if(nums.Length == 0)
            return 0;
        int max = int.MinValue;
        for(int i = 0; i < nums.Length; i++){
            int tempSum=0;
            for(int j =i; j < nums.Length; j++){
                tempSum += nums[j];
                if(tempSum > max)
                max = tempSum;
            }
        } 
        return max;
    }
}

看题目描述，可以使用divide and conquer（分而治之）思想去实现，时间复杂度会大幅度降低：这里先将解法贴出来，具体的分而治之（动态规划）的学习放到下一个文章中

static int maxCrossingSum(int[] arr, int l,
                                int m, int h)
        {
            // Include elements on left of mid.
            int sum = 0;
            int left_sum = int.MinValue;
            for (int i = m; i >= l; i--)
            {
                sum = sum + arr[i];
                if (sum > left_sum)
                    left_sum = sum;
            }

            // Include elements on right of mid
            sum = 0;
            int right_sum = int.MinValue; ;
            for (int i = m + 1; i <= h; i++)
            {
                sum = sum + arr[i];
                if (sum > right_sum)
                    right_sum = sum;
            }

            // Return sum of elements on left
            // and right of mid
            return left_sum + right_sum;
        }

        // Returns sum of maxium sum subarray 
        // in aa[l..h]
        static int maxSubArraySum(int[] arr, int l,
                                            int h)
        {

            // Base Case: Only one element
            if (l == h)
                return arr[l];

            // Find middle point
            int m = (l + h) / 2;

            /* Return maximum of following three 
            possible cases:
            a) Maximum subarray sum in left half
            b) Maximum subarray sum in right half
            c) Maximum subarray sum such that the 
            subarray crosses the midpoint */
            return Math.Max(Math.Max(maxSubArraySum(arr, l, m),
                                  maxSubArraySum(arr, m + 1, h)),
                                 maxCrossingSum(arr, l, m, h));
        }

        /* Driver program to test maxSubArraySum */
        public static void Main()
        {
            int[] arr = { -2, 3, 4, -5, 7 };
            int n = arr.Length;
            int max_sum = maxSubArraySum(arr, 0, n - 1);

            Console.Write("Maximum contiguous sum is " +
                                                max_sum);
            Console.ReadKey();
        }

分而治之的链接：算法学习分而治之思想

LeetCode Array Easy 53. Maximum Subarray 个人解法 和分治思想的学习

LeetCode Array Easy 53. Maximum Subarray 个人解法和分治思想的学习