[英]How to iterate over array of integers to find a sequence based on an O(N) solution?
我看到以下問題並試圖找到答案。
Question: Given a sequence of positive integers A and an integer T, return whether there is a *continuous sequence* of A that sums up to exactly T
Example
[23, 5, 4, 7, 2, 11], 20. Return True because 7 + 2 + 11 = 20
[1, 3, 5, 23, 2], 8. Return True because 3 + 5 = 8
[1, 3, 5, 23, 2], 7 Return False because no sequence in this array adds up to 7
Note: We are looking for an O(N) solution. There is an obvious O(N^2) solution which is a good starting point but is not the final solution we are looking for.
我對上述問題的回答是:
public class Tester {
public static void main(String[] args) {
int[] myArray = {23, 5, 4, 7, 2, 11};
System.out.println(isValid(myArray, 20));
}
public static boolean isValid(int[] array, int sum) {
int pointer = 0;
int temp = 0;
while (pointer < array.length)
{
for (int i = pointer; i < array.length; i++)
{
if (array[i] > sum)
break;
temp += array[i];
if (temp == sum)
return true;
else if (temp > sum)
break;
// otherwise continue
}
temp = 0;
pointer++;
}
return false;
}
}
我認為我的答案是 O(N^2),根據問題,這是不可接受的。 是否有基於 O(N) 的解決方案?
您實際上只需要循環一次,即 O(N)。
從索引 0 開始添加,一旦超過sum
就從數組的開頭開始刪除。 如果temp
低於sum
繼續循環。
public static boolean isValid(int[] array, int sum) {
int init = 0,temp = 0;
for (int i = 0; i < array.length; i++) {
temp += array[i];
while (temp > sum) {
temp -= array[init];
init++;
}
if (temp == sum)
return true;
}
return false;
}
您應該做的是有兩個索引(開始和停止),然后增加stop
直到總和達到要求(並返回true
)或更高。 然后你增加start
直到總和是所需的(並返回true
或以下。然后你重復這個直到你到達數組的末尾。你可以增量更新總和(當你增加stop
時添加元素,當你增加start
時減去元素) ). 這應該是 O(N)。
下面是一個例子:
public class t {
public static void main(String[] args) {
int[] myArray = {23, 5, 4, 7, 2, 11};
System.out.println(isValid(myArray, 20));
}
public static boolean isValid(int[] array, int sum) {
int start = 0;
int stop = 0;
int tsum = 0;
while( true )
{
if( tsum < sum )
{
if( stop >= array.length )
break;
tsum += array[stop];
stop++;
}
else if( tsum > sum )
{
tsum -= array[start];
start++;
}
else if( tsum == sum )
return true;
// System.out.println(start + " -- " + stop + " => " + tsum);
}
return false;
}
}
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