Java 窗口k算法的最大和子数组未正确返回起始索引
这是我的maxSum子数组代码 给定输入数组->[110,-4,3,6,7,11]和k=3,代码应给出[110,-4,3],因为这是具有最大和的子数组 我已经实现了这个算法。它适用于除一个输入之外的所有输入。我不知道出了什么问题Java 窗口k算法的最大和子数组未正确返回起始索引,java,arrays,algorithm,sub-array,Java,Arrays,Algorithm,Sub Array,这是我的maxSum子数组代码 给定输入数组->[110,-4,3,6,7,11]和k=3,代码应给出[110,-4,3],因为这是具有最大和的子数组 我已经实现了这个算法。它适用于除一个输入之外的所有输入。我不知道出了什么问题 public class MaxSumSizeK { private static int getMaxAvgSubarrayStartIndex(int input[], int k) { int n = i
public class MaxSumSizeK {
private static int getMaxAvgSubarrayStartIndex(int input[], int k)
{
int n = input.length;
if (k > n)
throw new IllegalArgumentException("k should be less than or equal to n" );
if(k == n) {
return 0;
}
int sum = input[0];
for (int i = 1; i < k; i++)
sum += input[i];
int maxSum = sum;
int maxSumIndex = 0;
for (int i = k; i < n; i++){
sum = sum - input[i-k] + input[i] ;
if (sum > maxSum){
maxSum = sum;
maxSumIndex = i-k;
}
}
return maxSumIndex+1;
}
public static void printMaxAvgSubarray(int[] input, int k) {
System.out.print("Maximum average subarray of length " + k + " is: " );
int index = getMaxAvgSubarrayStartIndex(input, k);
for(int i =0 ; i < k; i++) {
System.out.print(input[index++] + " " );
}
}
public static void main(String[] args) {
int[] input = {11, -8, 16, -7, 24, -2, 300};
int k = 3;
printMaxAvgSubarray(input, k);
System.out.println();
int[] input1 = {110, -8, 16, -7, 24, -2, 3};
printMaxAvgSubarray(input1, k);
System.out.println();
int[] input2 = {11, -8, 16, -7, 24, -2, 3};
printMaxAvgSubarray(input2, k);
System.out.println();
}
}
对于输入1,我看不到预期的答案应该是“110,-8,16”。我尝试将return语句改为“maxSumIndex”,而不是“maxSumIndex+1”。这会打断其他两个输入
请提供您的想法有两个错误:
maxSumIndex=i-k代码>应该是maxSumIndex=i-k+1代码>i-k-它是前一个数组的第一个索引,而不是当前的最大索引
返回maxSumIndex+1
应该是返回maxSumIndex代码>如果索引为0(即前3项为最大值),则返回1,这是错误的
int n = input.length;
if (k > n)
throw new IllegalArgumentException("k should be less than or equal to n" );
if (k < 1)
throw new IllegalArgumentException("k should be greater than 0" );
if(k == n) {
return 0;
}
int sum = 0;
for (int i = 0; i < k; i++)
sum += input[i];
int maxSum = sum;
int maxSumIndex = 0;
for (int j = 1; j < n-k; j++){
sum = 0;
for (int i = 0; i < k; i++){
sum = sum + (int) Math.abs(input[i+j]);
}
if (sum > maxSum){
maxSum = sum;
maxSumIndex = i-k;
}
}
return maxSumIndex;
int n=input.length;
如果(k>n)
抛出新的IllegalArgumentException(“k应小于或等于n”);
if(k<1)
抛出新的IllegalArgumentException(“k应大于0”);
如果(k==n){
返回0;
}
整数和=0;
for(int i=0;i最大总和){
最大和=和;
maxSumIndex=i-k;
}
}
返回maxSumIndex;
为什么[110,-4,3]是[110,-4,3,6,7,11]的最大和?您的期望看起来像是要打印出第一项。@NikolasCharalambidis可能是连续数字的唯一子集?所以你不能到处乱跳?但是,是的,我想知道这是数组中3个元素的最大和,110,-4,3,6,7,11
是110+11+7
,它是128
,所以我问..@nikolasharalalambidis:它是子数组,意味着它是连续的。110,11,7不是连续的您可以使用一个for循环而不是嵌套循环
int n = input.length;
if (k > n)
throw new IllegalArgumentException("k should be less than or equal to n" );
if (k < 1)
throw new IllegalArgumentException("k should be greater than 0" );
if(k == n) {
return 0;
}
int sum = 0;
for (int i = 0; i < k; i++)
sum += input[i];
int maxSum = sum;
int maxSumIndex = 0;
for (int j = 1; j < n-k; j++){
sum = 0;
for (int i = 0; i < k; i++){
sum = sum + (int) Math.abs(input[i+j]);
}
if (sum > maxSum){
maxSum = sum;
maxSumIndex = i-k;
}
}
return maxSumIndex;