java求最大子数组（分治算法）

最新推荐文章于 2021-12-16 21:38:15 发布

Lamb_quan

最新推荐文章于 2021-12-16 21:38:15 发布

阅读量817

点赞数

CC 4.0 BY-SA版权

分类专栏：算法文章标签：分治算法最大子数组

本文链接：https://blog.youkuaiyun.com/Lamb_quan/article/details/88394170

算法专栏收录该内容

4 篇文章

订阅专栏

本文介绍了一种求解最大子数组和的算法实现，通过递归方式寻找包含负数的最大子数组，给出具体代码示例及运行结果，展示了算法的有效性和效率。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

当一个数组有负数时，最大子数组才会有意义。

package shu.quan.demo;

/**
 * 求最大子串的值，并求出脚标。
 */
public class MaxSubsequence {
    static int ileft = 0, iright = 0;
    public static void main(String[] args){
        int[] arr = new int[]{1,5,-5,6,-13,4,8,-5,14};
        int res = maxSubsequence(arr);
        System.out.print("["+ileft+"，"+iright+"]"+res);

    }

    static int maxSubsequence(int[] arr){
        return maxSubsequence(arr,0,arr.length - 1);
    }

    static int maxSubsequence(int[] arr, int left, int right){
        if(left == right){
            return arr[left];
        }else {
            int center = (left + right) /2;

            int maxLeftSum = maxSubsequence(arr, left, center);
            int maxRightSum = maxSubsequence(arr, center + 1, right);
            int maxLeftSubsequenceSum = 0, maxLeft = 0;
            for (int i = center; i >= left; i--) {
                maxLeftSubsequenceSum += arr[i];
                if (maxLeftSubsequenceSum > maxLeft){
                    maxLeft = maxLeftSubsequenceSum;
                    ileft = i;
                }

            }

            int maxRightSubsequenceSum = 0, maxRight = 0;
            for (int i = center + 1; i <= right; i++) {
                maxRightSubsequenceSum += arr[i];
                if (maxRightSubsequenceSum > maxRight){
                    maxRight = maxRightSubsequenceSum;
                    iright = i;
                }

            }

            return Math.max(Math.max(maxLeftSum,maxLeft+maxRight),Math.max(maxRightSum,maxLeft+maxRight));

        }
    }
}

运行结果：

[5，8]21
Process finished with exit code 0