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

Posted 2020-10-31 C_supreme

Description

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