LeetCode 53. Maximum Subarray

53. Maximum Subarray

my program:

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int result = nums[0];
        int sum = 0;
        for (int i = 0; i<nums.size(); ++i)
        {
            sum = 0;
            for (int j = i; j<nums.size();++j)
            {
                sum += nums[j];
                if (sum > result)   result = sum;
            }
        }
        return result;
    }
};

Submission Result: Time Limit Exceeded

这个就是所谓的蛮力法吧…….%>_<%
时间复杂度是O(n²),当数据集越大,执行所需的时间就成平方数量级的增长,所以计算所需的时间超出了规定的时间,无法AC.

分治算法

首先将其分成两半A[l..m]和A[m+1..r]，其中m=(l+r)/2，并分别求递归求出这两半的最大子串和

中间middle部分是两个循环,分别找出左边和右面(包含m点)的最大子串和.

最后取left\right\middle中最大值

程序如下:

class Solution {
public:
    int solve(vector<int>& nums, int begin, int end)
    {
        if (begin == end) return nums[begin];
        int mid = (begin+end) / 2;
        int leftMax = solve(nums, begin, mid);
        int rightMax = solve(nums, mid+1, end);
        int midLeftMax = nums[mid];
        int sumMidLeft = nums[mid];
        for (int i = mid-1; i>=begin; --i)   
        {
            sumMidLeft += nums[i];
            midLeftMax = max(midLeftMax, sumMidLeft);
        }
        int midRightMax = nums[mid];
        int sumMidRight = nums[mid];
        for (int j = mid+1; j<=end; ++j)  
        {
            sumMidRight += nums[j];
            midRightMax = max(midRightMax, sumMidRight);
        }
        return max(max(leftMax, rightMax),midLeftMax+midRightMax-nums[mid]);
    }
    int maxSubArray(vector<int>& nums) {
      return solve(nums, 0, nums.size()-1);
    }
};

Simplest and fastest O(n) C++ solution:

Idea is very simple. Basically, keep adding each integer to the sequence until the sum drops below 0.
If sum is negative, then should reset the sequence.

class Solution {
public:
    int maxSubArray(int A[], int n) {
        int ans=A[0],i,j,sum=0;
        for(i=0;i<n;i++){
            sum+=A[i];
            ans=max(sum,ans);
            sum=max(sum,0);
        }
        return ans;
    }
};

此算法的时间复杂度是O(n),当数据非常大的时候,比起时间复杂度是O(n²)的算法好的不是一点.