[Leetcode]Maximum Subarray

最新推荐文章于 2021-10-21 15:55:14 发布

dyllanzhou

最新推荐文章于 2021-10-21 15:55:14 发布

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分类专栏： algorithm Array Divide and Conquer DP 文章标签： leetcode 算法

本文链接：https://blog.youkuaiyun.com/dyllanzhou/article/details/48654999

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本文介绍两种求解最大子数组和问题的算法：分治法和动态规划法。分治法通过递归将问题分解为子问题并求解，而动态规划法则采用一维状态转移方程逐步求解。两种方法均提供了高效解决方案。

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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.

class Solution {
public:
    /*algorithm  divide and conquer
      A[1..n] can divide to A[1..i-1] A[i] A[i+1..n]
      f(1..n) = max(f(1..i-1),f(i+1...n),union(f(1..i-1)+A[i]+f(i+1..n))
      time O(nlogn) space O(nlogn)
    */
    //[start,end)
    int maxSubArrayHelper(vector<int>&nums,int start,int end)
    {
        if((end - start) < 1)return INT_MIN;
        if((end-start) < 2)return nums[start];
        int mid = start + (end-start)/2;
        int left = maxSubArrayHelper(nums,start,mid);
        int right = maxSubArrayHelper(nums,mid+1,end);
        int lsum = 0,lsmax=0;
        for(int k=mid-1;k >= start;k--){
            lsum += nums[k];
            lsmax = max(lsmax,lsum);
        }
        int rsum = 0,rsmax=0;
        for(int k=mid+1;k < end;k++){
            rsum += nums[k];
            rsmax = max(rsum,rsmax);
        }
        return max(max(left,right),lsmax+nums[mid]+rsmax);
    }
    int maxSubArray(vector<int>& nums) {
        return maxSubArrayHelper(nums,0,nums.size());
    }
};

class Solution {
public:
    /*algorithm dp solution
     for array,A[1...n]
     f(1)=A[1];
     f(2)=max(f(1) + A[2],A[2])
     f(n) = max(f(n-1) + A[n],A[n])
     time O(n) space O(1)
    */
    int maxSubArray(vector<int>& nums) {
        int n = nums.size();
        int gmax = nums[0];
        int f0 = nums[0],f1;
        for(int i = 1;i < n;i++){
            f1 = max(f0+nums[i],nums[i]);
            f0 = f1;
            gmax = max(gmax,f0);
        }
        return gmax;
    }
};