使用分治策略实现求最大子数组——Java版

给定一个数组，找到一个具备最大和的连续子数组（子数组最少包含一个元素），返回其最大和。

示例：

输入: [-2,1,-3,4,-1,2,1,-5,4],
输出: 6
解释: 连续子数组 [4,-1,2,1] 的和最大，为 6。

 

public class 最大子数组 {

    public static void main(String[] args) {
        int[] res = getMaxArray(new int[]{-2,1,-3,4,-1,2,1,-5,4},0,8);
        for (int i = 0; i < res.length; i++) {
            System.out.println(res[i]);
        }
    }

    private static int[] getMaxArray(int[] A,int low,int high){
        if (low == high){
            int[] res = {low,high,A[low]};
            return res;
        }else {
            int mid = (int) Math.floor((low+high)/2);
            int[] left = getMaxArray(A,low,mid);
            int[] right = getMaxArray(A,mid+1,high);
            int[] cross = getCrossMaxArray(A,low,high,mid);
            if (left[1] >= right[1] && left[1] >= cross[1]){
                return left;
            } else if (right[2] >= left[2] && right[2] >= cross[2]){
                return right;
            } else{
                return cross;
            }
        }
    }

    private static int[] getCrossMaxArray(int[] A,int low,int high,int mid){
        int leftSum = 0;
        int sum1 = 0;
        int leftIndex = 0;
        for (int i = mid; i >= low; i--) {
            sum1 += A[i];
            if (sum1 > leftSum){
                leftSum = sum1;
                leftIndex = i;
            }
        }

        int rightSum = 0;
        int sum2 = 0;
        int rightIndex = 0;
        for (int i = mid+1; i < high; i++) {
            sum2 += A[i];
            if (sum2 > rightSum){
                rightSum = sum2;
                rightIndex = i;
            }
        }

        int res[] = {leftIndex,rightIndex,leftSum+rightSum};
        return res;
    }

}