题目描述
给定一个数组和滑动窗口的大小,请找出所有滑动窗口里的最大值。例如,如果输入数组{2, 3, 4, 2, 6, 2, 5, 1}及滑动窗口的大小3,那么一共存在6个滑动窗口,它们的最大值分别为{4, 4, 6, 6, 6, 5}。
算法分析
用一个双端队列,队列第一个位置保存当前窗口的最大值,当窗口滑动一次判断当前最大值是否过期;新增加的值从队尾开始比较,把所有比他小的值丢掉。
提交代码:
class Solution {
public:
vector<int> maxInWindows(const vector<int>& num, unsigned int size)
{
if (num.empty() || size < 1 || size > num.size())
return vector<int>();
deque<int> maxisium{0};
vector<int> result;
int i = 1;
for (; i < size; ++i)
{
while (!maxisium.empty() && num[i] > num[maxisium.back()])
maxisium.pop_back();
maxisium.push_back(i);
}
result.push_back(num[maxisium.front()]);
for (; i < num.size(); ++i)
{
if(!maxisium.empty() && (i - maxisium.front() + 1) > size)
maxisium.pop_front();
while (!maxisium.empty() && num[i] > num[maxisium.back()])
maxisium.pop_back();
maxisium.push_back(i);
result.push_back(num[maxisium.front()]);
}
return result;
}
};
测试代码:
// ====================测试代码====================
void Test(const char* testName, const vector<int>& num, unsigned int size, const vector<int>& expected)
{
if (testName != nullptr)
printf("%s begins: ", testName);
Solution s;
vector<int> result = s.maxInWindows(num, size);
vector<int>::const_iterator iterResult = result.begin();
vector<int>::const_iterator iterExpected = expected.begin();
while (iterResult < result.end() && iterExpected < expected.end())
{
if (*iterResult != *iterExpected)
break;
++iterResult;
++iterExpected;
}
if (iterResult == result.end() && iterExpected == expected.end())
printf("Passed.\n");
else
printf("FAILED.\n");
}
void Test1()
{
int num[] = { 2, 3, 4, 2, 6, 2, 5, 1 };
vector<int> vecNumbers(num, num + sizeof(num) / sizeof(int));
int expected[] = { 4, 4, 6, 6, 6, 5 };
vector<int> vecExpected(expected, expected + sizeof(expected) / sizeof(int));
unsigned int size = 3;
Test("Test1", vecNumbers, size, vecExpected);
}
void Test2()
{
int num[] = { 1, 3, -1, -3, 5, 3, 6, 7 };
vector<int> vecNumbers(num, num + sizeof(num) / sizeof(int));
int expected[] = { 3, 3, 5, 5, 6, 7 };
vector<int> vecExpected(expected, expected + sizeof(expected) / sizeof(int));
unsigned int size = 3;
Test("Test2", vecNumbers, size, vecExpected);
}
// 输入数组单调递增
void Test3()
{
int num[] = { 1, 3, 5, 7, 9, 11, 13, 15 };
vector<int> vecNumbers(num, num + sizeof(num) / sizeof(int));
int expected[] = { 7, 9, 11, 13, 15 };
vector<int> vecExpected(expected, expected + sizeof(expected) / sizeof(int));
unsigned int size = 4;
Test("Test3", vecNumbers, size, vecExpected);
}
// 输入数组单调递减
void Test4()
{
int num[] = { 16, 14, 12, 10, 8, 6, 4 };
vector<int> vecNumbers(num, num + sizeof(num) / sizeof(int));
int expected[] = { 16, 14, 12 };
vector<int> vecExpected(expected, expected + sizeof(expected) / sizeof(int));
unsigned int size = 5;
Test("Test4", vecNumbers, size, vecExpected);
}
// 滑动窗口的大小为1
void Test5()
{
int num[] = { 10, 14, 12, 11 };
vector<int> vecNumbers(num, num + sizeof(num) / sizeof(int));
int expected[] = { 10, 14, 12, 11 };
vector<int> vecExpected(expected, expected + sizeof(expected) / sizeof(int));
unsigned int size = 1;
Test("Test5", vecNumbers, size, vecExpected);
}
// 滑动窗口的大小等于数组的长度
void Test6()
{
int num[] = { 10, 14, 12, 11 };
vector<int> vecNumbers(num, num + sizeof(num) / sizeof(int));
int expected[] = { 14 };
vector<int> vecExpected(expected, expected + sizeof(expected) / sizeof(int));
unsigned int size = 4;
Test("Test6", vecNumbers, size, vecExpected);
}
// 滑动窗口的大小为0
void Test7()
{
int num[] = { 10, 14, 12, 11 };
vector<int> vecNumbers(num, num + sizeof(num) / sizeof(int));
vector<int> vecExpected;
unsigned int size = 0;
Test("Test7", vecNumbers, size, vecExpected);
}
// 滑动窗口的大小大于输入数组的长度
void Test8()
{
int num[] = { 10, 14, 12, 11 };
vector<int> vecNumbers(num, num + sizeof(num) / sizeof(int));
vector<int> vecExpected;
unsigned int size = 5;
Test("Test8", vecNumbers, size, vecExpected);
}
// 输入数组为空
void Test9()
{
vector<int> vecNumbers;
vector<int> vecExpected;
unsigned int size = 5;
Test("Test9", vecNumbers, size, vecExpected);
}
int main(int argc, char* argv[])
{
Test1();
Test2();
Test3();
Test4();
Test5();
Test6();
Test7();
Test8();
Test9();
return 0;
}