我有一个数字数组，我想算出两个重复的唯一数字组成的连续子数组的最大长度。
例如，[2、3、4、3、2、2、4]将返回3，因为[3、2、2]的长度为3。
[2, 4, 2, 5, 1, 5, 4, 2] would return 3.
[7, 8, 7, 8, 7] would return 5.
编辑：我考虑过一个O（n ^ 2）解决方案，其中我从数组中的每个值开始并进行迭代，直到看到第三个唯一值为止。
for item in array:
iterate until third unique element
if length of this iteration is greater than existing max, update the max length
return maxlength
但是，我认为这不是一个有效的解决方案。
答案 0 :(得分：2)
可以完成O（n）。该代码在python3中。 o和t分别是1和2。 m是最大值，c是当前计数变量。
a = [7, 8, 7, 8, 7]
m = -1
o = a[0]
t = a[1]
# in the beginning one and two are the first 2 numbers
c = 0
index = 0
for i in a:
if i == o or i == t:
# if current element is either one or two current count is increased
c += 1
else:
# if current element is neither one nor two then they are updated accordingly and max is updated
o = a[index - 1]
t = a[index]
m = max(m, c)
c = 2
index += 1
m = max(m, c)
print(m)
答案 1 :(得分：0)
我们可以使用两个指针技术来解决O（n）运行时复杂性的问题。这两个指针，例如startPtr和endPtr将表示数组中的范围。我们将以不超过2个唯一编号的方式维护此范围[startPtr, endPtr]。我们可以通过跟踪2个唯一编号的位置来做到这一点。我的C ++工具如下：
int main()
{
int array[] = {1,2,3,3,2,3,2,3,2,2,2,1,3,4};
int startPtr = 0;
int endPtr = 0;
// denotes the size of the array
int size= sizeof(array)/sizeof(array[0]);
// contain last position of unique number 1 in the range [startPtr, endPtr]
int uniqueNumPos1 = -1; // -1 value represents it is not set yet
// contain last position of unique number 2 in the range [startPtr, endPtr]
int uniqueNumPos2 = -1; // -1 value represents it is not set yet
// contains length of maximum continuous subarray with 2 unique numbers
int ans = 0;
while(endPtr < size) {
if(uniqueNumPos1 == -1 || array[endPtr] == array[uniqueNumPos1]) {
uniqueNumPos1 = endPtr;
}
else {
if(uniqueNumPos2 == -1 || array[endPtr] == array[uniqueNumPos2]) {
uniqueNumPos2 = endPtr;
}
else {
// for this new third unique number update startPtr with min(uniqueNumPos1, uniqueNumPos2) + 1
// to ensure [startPtr, endPtr] does not contain more that two unique
startPtr = min(uniqueNumPos1, uniqueNumPos2) + 1;
// update uniqueNumPos1 and uniqueNumPos2
uniqueNumPos1 = endPtr -1;
uniqueNumPos2 = endPtr;
}
}
// this conditon is to ensure the range contain exactly two unique number
// if you are looking for the range containing less than or equal to two unique number, then you can omit this condition
if (uniqueNumPos1 != -1 && uniqueNumPos2 !=-1) {
ans = max( ans, endPtr - startPtr + 1);
}
endPtr++;
}
printf("%d\n", ans);
}
感谢 @MBo 指出错误。
答案 2 :(得分：0)
import java.util.Arrays;
import static java.lang.System.out;
class TestCase{
int[] test;
int answer;
TestCase(int[] test,int answer){
this.test = test;
this.answer = answer;
}
}
public class Solution {
public static void main(String[] args) {
TestCase[] tests = {
new TestCase(new int[]{2, 3, 4, 3, 2, 2, 4},3),
new TestCase(new int[]{2, 3, 3, 3, 3, 4, 3, 3, 2, 2, 4},7),
new TestCase(new int[]{1,2,3,3,4,2,3,2,3,2,2,2,1,3,4},7),
new TestCase(new int[]{2, 7, 8, 7, 8, 7},5),
new TestCase(new int[]{-1,2,2,2,2,2,2,2,2,2,2,-1,-1,4},13),
new TestCase(new int[]{1,2,3,4,5,6,7,7},3),
new TestCase(new int[]{0,0,0,0,0},0),
new TestCase(new int[]{0,0,0,2,2,2,1,1,1,1},7),
new TestCase(new int[]{},0)
};
for(int i=0;i<tests.length;++i){
int ans = maxContiguousArrayWith2UniqueElements(tests[i].test);
out.println(Arrays.toString(tests[i].test));
out.println("Expected: " + tests[i].answer);
out.println("Returned: " + ans);
out.println("Result: " + (tests[i].answer == ans ? "ok" : "not ok"));
out.println();
}
}
private static int maxContiguousArrayWith2UniqueElements(int[] A){
if(A == null || A.length <= 1) return 0;
int max_subarray = 0;
int first_number = A[0],second_number = A[0];
int start_index = 0,same_element_run_length = 1;
for(int i=1;i<A.length;++i){
if(A[i] != A[i-1]){
if(first_number == second_number){
second_number = A[i];
}else{
if(A[i] != first_number && A[i] != second_number){
max_subarray = Math.max(max_subarray,i - start_index);
start_index = i - same_element_run_length;
first_number = A[i-1];
second_number = A[i];
}
}
same_element_run_length = 1;
}else{
same_element_run_length++;
}
}
return first_number == second_number ? max_subarray : Math.max(max_subarray,A.length - start_index);
}
}
输出：
[2, 3, 4, 3, 2, 2, 4]
Expected: 3
Returned: 3
Result: ok
[2, 3, 3, 3, 3, 4, 3, 3, 2, 2, 4]
Expected: 7
Returned: 7
Result: ok
[1, 2, 3, 3, 4, 2, 3, 2, 3, 2, 2, 2, 1, 3, 4]
Expected: 7
Returned: 7
Result: ok
[2, 7, 8, 7, 8, 7]
Expected: 5
Returned: 5
Result: ok
[-1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, -1, -1, 4]
Expected: 13
Returned: 13
Result: ok
[1, 2, 3, 4, 5, 6, 7, 7]
Expected: 3
Returned: 3
Result: ok
[0, 0, 0, 0, 0]
Expected: 0
Returned: 0
Result: ok
[0, 0, 0, 2, 2, 2, 1, 1, 1, 1]
Expected: 7
Returned: 7
Result: ok
[]
Expected: 0
Returned: 0
Result: ok
算法：