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Java 分治最大连续子阵（MCS）问题

java algorithm

Java 分治最大连续子阵（MCS）问题,java,algorithm,divide-and-conquer,sub-array,Java,Algorithm,Divide And Conquer,Sub Array,我想为给定的整数输入数组找到一个非空、连续的子数组，它可以有重复的值。我尝试了分而治之的方法来查找数组的最大连续子数组，它会返回与预期不同的结果。请在下面查找代码 private static int maxSumRec(int[] a, int low, int high) { int leftSum = 0, rightSum = 0; int sum = 0; if (low == high) { // Base case return a[lo

我想为给定的整数输入数组找到一个非空、连续的子数组，它可以有重复的值。我尝试了分而治之的方法来查找数组的最大连续子数组，它会返回与预期不同的结果。请在下面查找代码

private static int maxSumRec(int[] a, int low, int high) {
    int leftSum = 0, rightSum = 0;
    int sum = 0; 

    if (low == high) { // Base case
        return a[low]; 
    }

    int mid = (low + high) >> 1; // (low + high) / 2
    int maxLeftSum = maxSumRec(a, low, mid);
    int maxRightSum = maxSumRec(a, mid + 1, high);

    //max-crossing-subarray
    for (int i = mid; i >= low; i--) {
        sum += a[i];
        if (sum > leftSum) {
            leftSum = sum;
        }
    }
    sum = 0;
    for (int i = mid + 1; i <= high; i++) {
        sum += a[i];
        if (sum > rightSum) {
            rightSum = sum;
        }
    }
    return max3(maxLeftSum, maxRightSum, (leftSum + rightSum));
}

private static int max3(int a, int b, int c) {
    return a > b ? (a > c ? a : c) : (b > c ? b : c);
}

public static void main(String[] args) {


    //INPUT
    int a[] = {
        -5, 71, 23, 41, 34, 1, 3, 6, 2, 91, 312, 42, 31, 67, 12, 10, 18, 56, 90, 21, 45, 47, 89, 1999999990,
        78, -7, 76, 75, 74, 73, 72, 80, 24, 25, 61, 69, 84, 0, -1, 145, 1902, 200, 199, 198, 197, 196, 195, 194,
        78, 77, 76, 75, 74, 73, 72, 80, 24, 25, 61, 69, 84, 0, -1, 145, 1902, 200, 199, 198, 197, 196, 195, 194,
        5, 71, 23, 41, 34, 1, 3, 6, 2, 91, 312, 42, 31, 67, 12, 10, 18, 56, 90, 21, 45, 47, 89, 1999999990
    };

    int maxSum = maxSumRec(a, 0, a.length - 1);
    System.out.println("Max sum is " + maxSum);
}

private static int-maxSumRec（int[]a，int-low，int-high）{
int leftSum=0，rightSum=0；
整数和=0；
如果（低==高）{//基本情况
返回一个[低]；
}
int mid=（低+高）>>1；//（低+高）/2
int maxLeftSum=maxSumRec（a、低、中）；
int maxRightSum=maxSumRec（a，中+1，高）；
//最大交叉子阵
对于（int i=mid；i>=low；i--）{
总和+=a[i]；
如果（总和>左总和）{
左和=和；
}
}
总和=0；
对于（整数i=mid+1；i右和）{
rightSum=总和；
}
}
返回max3（maxLeftSum，maxRightSum，（leftSum+rightSum））；
}
私有静态intmax3（inta，intb，intc）{
返回a>b？（a>c？a:c）：（b>c？b:c）；
}
公共静态void main（字符串[]args）{
//输入
INTA[]={
-5, 71, 23, 41, 34, 1, 3, 6, 2, 91, 312, 42, 31, 67, 12, 10, 18, 56, 90, 21, 45, 47, 89, 1999999990,
78, -7, 76, 75, 74, 73, 72, 80, 24, 25, 61, 69, 84, 0, -1, 145, 1902, 200, 199, 198, 197, 196, 195, 194,
78, 77, 76, 75, 74, 73, 72, 80, 24, 25, 61, 69, 84, 0, -1, 145, 1902, 200, 199, 198, 197, 196, 195, 194,
5, 71, 23, 41, 34, 1, 3, 6, 2, 91, 312, 42, 31, 67, 12, 10, 18, 56, 90, 21, 45, 47, 89, 1999999990
};
int maxSum=maxSumRec（a，0，a.length-1）；
System.out.println（“最大和为”+maxSum）；
}
此代码将结果返回为2000005400。MCS的非递归版本返回一个不同的结果，即2000010721及其在{1-94}中找到的结果。 我想不出原因。如果代码中有错误，请告诉我

从1到95的总和（即：4000010711）实际上大于从1到94的总和

您的
整数
太长

您需要使用
long
来获得正确的结果

注意：

public class Sample5 {
        private static long maxSumRec(int[] a, int low, int high) {
            long  leftSum = 0, rightSum = 0;
            long  sum = 0; 

            if (low == high) { // Base case
                return a[low]; 
            }

            int mid = (low + high)/2; // (low + high) / 2
            long maxLeftSum = maxSumRec(a, low, mid);
            long maxRightSum = maxSumRec(a, mid + 1, high);

            //max-crossing-subarray
            for (int i = mid; i >= low; i--) {
                sum += a[i];
                if (sum > leftSum) {
                    leftSum = sum;
                }
            }
            sum = 0;
            for (int i = mid + 1; i <= high; i++) {
                sum += a[i];
                if (sum > rightSum) {
                    rightSum = sum;
                }
            }
            System.out.println("final left sum "+leftSum);
            System.out.println("final right sum "+rightSum);
            System.out.println("leftSum+rightSUM:"+(leftSum + rightSum));
            return max3(maxLeftSum, maxRightSum, (leftSum + rightSum));
        }

        private static long max3(long a, long b, long c) {
            return a > b ? (a > c ? a : c) : (b > c ? b : c);
        }

        private static int sum(int[] a,int i,int j){
            int r=0;
            for(int k=i;k<=j;k++){
                r+=a[k];
            }
            return r;
        }
        public static void main(String[] args) {


            //INPUT
            int a[] = {
                -5, 71, 23, 41, 34, 1, 3, 6, 2, 91, 312, 42, 31, 67, 12, 10, 18, 56, 90, 21, 45, 47, 89, 1999999990,
                78, -7, 76, 75, 74, 73, 72, 80, 24, 25, 61, 69, 84, 0, -1, 145, 1902, 200, 199, 198, 197, 196, 195, 194,
                78, 77, 76, 75, 74, 73, 72, 80, 24, 25, 61, 69, 84, 0, -1, 145, 1902, 200, 199, 198, 197, 196, 195, 194,
                5, 71, 23, 41, 34, 1, 3, 6, 2, 91, 312, 42, 31, 67, 12, 10, 18, 56, 90, 21, 45, 47, 89, 1999999990
            };

            long maxSum = maxSumRec(a, 0, a.length-1);
            System.out.println("Max sum is " + maxSum);

            //WITH INTS
            System.out.println("with ints, the sum 1 to 94 is " + sum(a,1,94));
            System.out.println("with ints, the sum 1 to 95 is " + sum(a,1,95));
        }


}

final left sum 2000005311
final right sum 2000005400
leftSum+rightSUM:4000010711
Max sum is 4000010711
with ints, the sum 1 to 94 is 2000010721
with ints, the sum 1 to 95 is -294956585
整数介于-2147483648和2147483647之间

int test=2147483647;
System.out.println(test);
System.out.println(test+1);
您将获得：

2147483647
-2147483648
试试这个：

public class Sample5 {
        private static long maxSumRec(int[] a, int low, int high) {
            long  leftSum = 0, rightSum = 0;
            long  sum = 0; 

            if (low == high) { // Base case
                return a[low]; 
            }

            int mid = (low + high)/2; // (low + high) / 2
            long maxLeftSum = maxSumRec(a, low, mid);
            long maxRightSum = maxSumRec(a, mid + 1, high);

            //max-crossing-subarray
            for (int i = mid; i >= low; i--) {
                sum += a[i];
                if (sum > leftSum) {
                    leftSum = sum;
                }
            }
            sum = 0;
            for (int i = mid + 1; i <= high; i++) {
                sum += a[i];
                if (sum > rightSum) {
                    rightSum = sum;
                }
            }
            System.out.println("final left sum "+leftSum);
            System.out.println("final right sum "+rightSum);
            System.out.println("leftSum+rightSUM:"+(leftSum + rightSum));
            return max3(maxLeftSum, maxRightSum, (leftSum + rightSum));
        }

        private static long max3(long a, long b, long c) {
            return a > b ? (a > c ? a : c) : (b > c ? b : c);
        }

        private static int sum(int[] a,int i,int j){
            int r=0;
            for(int k=i;k<=j;k++){
                r+=a[k];
            }
            return r;
        }
        public static void main(String[] args) {


            //INPUT
            int a[] = {
                -5, 71, 23, 41, 34, 1, 3, 6, 2, 91, 312, 42, 31, 67, 12, 10, 18, 56, 90, 21, 45, 47, 89, 1999999990,
                78, -7, 76, 75, 74, 73, 72, 80, 24, 25, 61, 69, 84, 0, -1, 145, 1902, 200, 199, 198, 197, 196, 195, 194,
                78, 77, 76, 75, 74, 73, 72, 80, 24, 25, 61, 69, 84, 0, -1, 145, 1902, 200, 199, 198, 197, 196, 195, 194,
                5, 71, 23, 41, 34, 1, 3, 6, 2, 91, 312, 42, 31, 67, 12, 10, 18, 56, 90, 21, 45, 47, 89, 1999999990
            };

            long maxSum = maxSumRec(a, 0, a.length-1);
            System.out.println("Max sum is " + maxSum);

            //WITH INTS
            System.out.println("with ints, the sum 1 to 94 is " + sum(a,1,94));
            System.out.println("with ints, the sum 1 to 95 is " + sum(a,1,95));
        }


}

final left sum 2000005311
final right sum 2000005400
leftSum+rightSUM:4000010711
Max sum is 4000010711
with ints, the sum 1 to 94 is 2000010721
with ints, the sum 1 to 95 is -294956585
没有足够的分数添加评论，因此创建了快速评论的答案：

Ricky的答案在大多数情况下都有效，但对于[-2，-1]这样的数组，它就不起作用了。只需添加：

leftSum=a[mid]；rightSum=a[mid+1]

请详细说明您试图解决的问题（Wikipedia链接就足够了），并提供示例输入和输出。此代码如“算法简介”[CLRS]中所述。对于我在代码中介绍的输入，算法似乎会中断，除非代码中有错误。您的输入样本不是集合，[元素重复]最大子数组是什么意思？你是否在寻找i，j，使得Sigma（a[k]）[对于每个i为什么不是[3,2,1，-4,5,2，-1,3]的和是11？