最大子段和問題 暴力窮舉法 分治算法 動態規劃

暴力窮舉法：

public class 最大子段和問題_暴力枚舉 {
    public static int MaxSubSunm(int a[])
    {
        int res=0;
        for(int i=0;i<a.length;i++)
        {
            int sum=0;
            for(int j=i;j<a.length;j++)
            {
                sum+=a[j];
                if(sum>res)  res=sum;
            }
        }
        return res;
    }

    public static void main(String[] args)
    {
        int a[]=new int[]{-1,2,4,7,6,-4,-7,5,-1};
       System.out.println(MaxSubSunm(a));
    }
}

---

分治：

/**
 * mid=(left+right)/2
 * 分成三部分：
 * 1.a[left,mid]
 * 2.a[mid+1,right]
 * 3.a[i][j],left<=i<=mid,mid+1<=j<=right
 */
public class 最大子段和問題_分治 {
    public static int MaxSubSum(int[] a,int left,int right)
    {
        int sum=0;
        if(left==right) sum=a[left]>0?a[left]:0;
        else{
            int mid=(left+right)/2;
            int leftsum=MaxSubSum(a,left,mid);
            int rightsum=MaxSubSum(a,mid+1,right);

            //接下來考慮橫跨mid兩邊的情況
            int s1=0,lefts=0;
            for(int i=mid;i>=left;i--)
            {
                lefts+=a[i];
                if (lefts>s1) s1=lefts;
            }
            int s2=0,rights=0;
            for(int j=mid+1;j<=right;j++)
            {
                rights+=a[j];
                if(rights>s2) s2=rights;
            }
            sum=s1+s2;
        }
        return sum;
    }

    public static void main(String[] args)
    {
        int a[]=new int[]{-1,2,4,7,6,-4,-7,5,-1};
        System.out.print(MaxSubSum(a,0,a.length-1));
    }
}

---

動態規劃：

bj=max{bj-1+aj,aj},1<=j<=n

時間複雜O(n)。

/**
 * bj=max{bj-1+aj,aj},1<=j<=n
 */
public class 最大子段和問題_動態規劃 {
    public static int MaxSubSum(int a[])
    {
        int temp=0,sum=0;
       for(int i=0;i<a.length;i++)
       {
           temp=Math.max(a[i],temp+a[i]);
           if(temp>sum) sum=temp;
       }
       return sum;
    }
    public static void main(String[] args)
    {
        int a[]=new int[]{-1,2,4,7,6,-4,-7,5,-1};
        System.out.print(MaxSubSum(a));
    }
}