最大子段和最大子段积java实现

package edu.pku.ss.hlj;
public class MaxSubSumRec   
{  
	/**
	 * 分治法求最大子段和
	 * @param a
	 * @param left
	 * @param right
	 * @return
	 */
    private static int maxSumRec(int[] a, int left, int right)  
    {  
        if (left == right)  
        {  
            if (a[left] > 0)  
            {  
            	System.out.print(a[left]+" ");
                return a[left];
            }  
            else  
            {  
                return 0;  
            }  
        }  
  
        int center = (left + right) / 2;  
        int maxLeftSum = maxSumRec(a, left, center);  
        int maxRightSum = maxSumRec(a, center + 1, right);  
  
        int maxLeftBorderSum = 0;  
        int leftBorderSum = 0;  
        for (int i = center; i >= left; i--)  
        {  
            leftBorderSum += a[i];  
            if (leftBorderSum > maxLeftBorderSum)  
            {  
                maxLeftBorderSum = leftBorderSum;  
            }  
        }  
  
        int maxRightBorderSum = 0;  
        int rightBorderSum = 0;  
        for (int i = center + 1; i <= right; i++)  
        {  
            rightBorderSum += a[i];  
            if (rightBorderSum > maxRightBorderSum)  
            {  
                maxRightBorderSum = rightBorderSum;  
            }  
        }  
  
        return max(maxLeftSum, maxRightSum, maxRightBorderSum + maxLeftBorderSum);  
    }  
  
    //入口程序  
    public static int maxSubSum(int[] a)  
    {  
        return maxSumRec(a, 0, a.length - 1);  
    }  
  
    //求三数中的最大值  
    private static int max(int a, int b, int c)  
    {  
        int max = a > b ? a : b;  
        max = c > max ? c : max;  
  
        return max;  
    }
    private static int max(int a,int b){
    	int max = a>b?a:b;
    	return max;
    }
    private static int min(int a,int b,int c){
    	int min = a<b?a:b;
    	min = min<c?min:c;
    	return min;
    }
    /**
     * 动态规划方法一
     * @param a
     * @return
     */
    public static int maxSubSum2(int[] a)  
    {  
        int maxSum = 0;  
        int thisSum = 0;  
  
        for (int j = 0; j < a.length; j++)  
        {  
            thisSum += a[j];  
            if (thisSum > maxSum)  
            {  
                maxSum = thisSum;  
            }  
            else if (thisSum < 0)  
            {  
                thisSum = 0;  
            }  
        }  
  
        return maxSum;  
    }
    /**
     * 动态规划方法二
     * @param a
     * @return
     */
    public static int maxSubSum3(int[] a){
    	int maxSum = 0;
    	int thisSum = 0;//记录前一个最大是子段和C[i]=max{c[i-1]+a[i],a[i]}
    	for(int i=0;i<a.length;i++){
    		if(thisSum >0){
    			thisSum += a[i];
    		}else{
    			thisSum = a[i];
    		}
    		if(thisSum > maxSum){
    			System.out.print(a[i] + " ");
    			maxSum = thisSum;
    		}
    	}
    	return maxSum;
    }
    /**
     * 动态规划方法三
     * @param a
     * @return
     */
    public static int maxS(int[] a){
    	int[] max = new int[25];
    	max[0] = a[0];
    	int maxSum = max[0];
    	for(int i=1;i<a.length;i++){
    		max[i] = max(max[i-1]+a[i],a[i]);
    		if(maxSum < max[i]){
    			maxSum = max[i];
    		}
    	}
    	return maxSum;
    }
    
    /**
     * 最大子段积（动态规划）
     * @param a
     */
    public static void multip(int[] a){
    	int[] min = new int[256];//记录到目前位置为止，最小的负数
    	int[] max = new int[256];//记录到目前位置为止，最大的正数
    	max[0] = a[0];//设置为第一个数值
    	min[0] = a[0];
    	int max_val = max[0];//记录最大乘积
    	for(int i=1;i<a.length;i++){
    		max[i] = max(max[i-1]*a[i],min[i-1]*a[i],a[i]);
    		min[i] = min(max[i-1]*a[i],min[i-1]*a[i],a[i]);//a[i]正负数都在这里了
    		if(max_val < max[i]){
    			max_val = max[i];
    		}
    	}
    	if(max_val < 0){
    		System.out.println(-1);
    	}else{
    		System.out.println(max_val);
    	}
    }
    public static void main(String[] args)   
    {  
        int[] a = {1,-2, 11, -4, 13, -5, -2}; 
        int[] b = {1, -3,-2,-2, -3,0,8,5};
//        System.out.println(MaxSubSumRec.maxSubSum(a));  
//        System.out.println(maxSubSum3(a));
        System.out.println(maxS(b));
//        multip(b);
    }  
}