LeetCode（53） Maximum Subarray

题目

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.

click to show more practice.

More practice:

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

分析

最大子序列和的问题，这道题我写出的是O(n)的算法，属于简单的动态规划，根据题目后面的more practice说明该题目还有更优的分治法解决思路。

AC代码-动态规划

class Solution {
public:
    int maxSubArray(vector<int>& nums) {

        if (nums.empty())
            return 0;

        //求数组的长度
        int len = nums.size();

        //将最大和赋值为首元素值，temp记录临时子序列和
        int maxSum = nums[0], temp = 0;
        for (int i = 0; i < len; i++)
        {
            temp += nums[i];

            //若元素和大于当前最大和
            if(temp > maxSum)
            {
                maxSum = temp;
            }//else

            //若子系列和为非正数，则从下一个元素重新记录
            if (temp <= 0)
            {
                temp = 0;
            }

        }//for

        return maxSum;
    }
};

AC代码-分治法

class Solution {
public:
    int maxSubArray(vector<int>& nums) {

        if (nums.empty())
            return 0;

        //求数组的长度
        int len = nums.size();

        return Divide(nums , 0 , len-1);
    }

    //分治法
    int Divide(const vector<int> &nums, int lhs, int rhs)
    {
        if (lhs == rhs)
            return nums[lhs];

        int mid = (lhs + rhs) / 2;
        int leftMaxSum = Divide(nums, lhs, mid);
        int rightMaxSum = Divide(nums, mid + 1, rhs);

        int lsum = INT_MIN;
        int rsum = INT_MIN;

        int temp = 0;
        for (int i = mid; i >= lhs; i--)
        {
            temp += nums[i];
            if (temp > lsum)
                lsum = temp;
        }

        temp = 0;
        for (int i = mid + 1; i <= rhs; i++)
        {
            temp += nums[i];
            if (temp > rsum)
                rsum = temp;
        }

        //跨越中点的最大子序列和
        temp = lsum + rsum;

        return std::max(temp, std::max(leftMaxSum, rightMaxSum));
    }
};

GitHub测试程序源码