给你一个整数数组 nums，有一个大小为 k 的滑动窗口从数组的最左侧移动到数组的最右侧。你只可以看到在滑动窗口内的 k 个数字。滑动窗口每次只向右移动一位。
返回 滑动窗口中的最大值 。
Example 1输入：nums = [1,3,-1,-3,5,3,6,7], k = 3 输出：[3,3,5,5,6,7] 解释： 滑动窗口的位置 最大值 --------------- ----- [1 3 -1] -3 5 3 6 7 3 1 [3 -1 -3] 5 3 6 7 3 1 3 [-1 -3 5] 3 6 7 5 1 3 -1 [-3 5 3] 6 7 5 1 3 -1 -3 [5 3 6] 7 6 1 3 -1 -3 5 [3 6 7] 7Example 2
输入：nums = [1], k = 1 输出：[1]
Constraints:
- 1 <= nums.length <= 10^5
- -10^4 <= nums[i] <= 10^4
- 1 <= k <= nums.length
class Solution {
public:
vector maxSlidingWindow(vector &nums, int k) {
deque q; // idx
auto push = [&](int idx) {
if (!q.empty() && q.front() == idx - k) q.pop_front();
while (!q.empty() && nums[q.back()] < nums[idx]) q.pop_back();
q.push_back(idx);
};
vector result;
for (int i = 0; i < nums.size(); ++i) {
push(i);
if (i + 1 >= k) result.push_back(nums[q.front()]);
}
return result;
}
};
2. 数组中的 k-diff 数对
给你一个整数数组 nums 和一个整数 k，请你在数组中找出 不同的 k-diff 数对，并返回不同的 k-diff 数对 的数目。
k-diff 数对定义为一个整数对 (nums[i], nums[j]) ，并满足下述全部条件：
- 0 <= i, j < nums.length
- i != j
- nums[i] - nums[j] == k
注意，|val| 表示 val 的绝对值。
Example 1输入：nums = [3, 1, 4, 1, 5], k = 2 输出：2 解释：数组中有两个 2-diff 数对, (1, 3) 和 (3, 5)。 尽管数组中有两个 1 ，但我们只应返回不同的数对的数量。Example 2
输入：nums = [1, 2, 3, 4, 5], k = 1 输出：4 解释：数组中有四个 1-diff 数对, (1, 2), (2, 3), (3, 4) 和 (4, 5) 。
Constraints:
- 1 <= nums.length <= 10^4
- -10^7 <= nums[i] <= 10^7
- 0 <= k <= 10^7
class Solution {
public:
int findPairs(vector &nums, int k) {
unordered_set uset, result;
for (int x:nums) {
if (uset.count(x - k)) result.insert(x - k);
if (uset.count(x + k)) result.insert(x);
uset.insert(x);
}
return result.size();
}
};
3. 下一个排列
A permutation of an array of integers is an arrangement of its members into a sequence or linear order.
- For example, for arr = [1,2,3], the following are considered permutations of arr: [1,2,3], [1,3,2], [3,1,2], [2,3,1].
The next permutation of an array of integers is the next lexicographically greater permutation of its integer. More formally, if all the permutations of the array are sorted in one container according to their lexicographical order, then the next permutation of that array is the permutation that follows it in the sorted container. If such arrangement is not possible, the array must be rearranged as the lowest possible order (i.e., sorted in ascending order).
- For example, the next permutation of arr = [1,2,3] is [1,3,2].
- Similarly, the next permutation of arr = [2,3,1] is [3,1,2].
- While the next permutation of arr = [3,2,1] is [1,2,3] because [3,2,1] does not have a lexicographical larger rearrangement.
Given an array of integers nums, find the next permutation of nums.
The replacement must be in place and use only constant extra memory.
Example 1Input: nums = [1,2,3] Output: [1,3,2]Example 2
Input: nums = [3,2,1] Output: [1,2,3]
Constraints:
- 1 <= nums.length <= 100
- 0 <= nums[i] <= 100
class Solution {
public:
void nextPermutation(vector &nums) {
int i = nums.size();
while (--i > 0 && nums[i - 1] >= nums[i]);
int idx_1 = i - 1, idx_2 = i;
if (i > 0) {
for (int j = i; j < nums.size(); ++j) {
if (nums[j] > nums[idx_1] && nums[j] < nums[idx_2]) idx_2 = j;
}
swap(nums[idx_1], nums[idx_2]);
}
sort(nums.begin() + i, nums.end());
}
};
代码 2 [O(n)]
代码1对于idx_1之后的所有元素使用快速排序，然而，只要保证idx_2是满足交换条件的最后一个元素，进行交换操作后，idx_1之后的元素递减，因此只需倒序即可。
class Solution {
public:
void nextPermutation(vector &nums) {
int i = nums.size();
while (--i > 0 && nums[i - 1] >= nums[i]);
int idx_1 = i - 1, idx_2 = i;
if (i > 0) {
for (int j = i; j < nums.size(); ++j) {
if (nums[j] > nums[idx_1] && nums[j] <= nums[idx_2]) idx_2 = j; // 此处与代码1不同
}
swap(nums[idx_1], nums[idx_2]);
}
int j = nums.size() - 1;
while (i < j) swap(nums[i++], nums[j--]);
}
};
4. 搜索旋转排序数组
There is an integer array nums sorted in ascending order (with distinct values).
Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,5,6,7] might be rotated at pivot index 3 and become [4,5,6,7,0,1,2].
Given the array nums after the possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.
You must write an algorithm with O(log n) runtime complexity.
Example 1Input: nums = [4,5,6,7,0,1,2], target = 0 Output: 4Example 2
Input: nums = [4,5,6,7,0,1,2], target = 3 Output: -1
Constraints:
- 1 <= nums.length <= 5000
- -10^4 <= nums[i] <= 10^4
- All values of nums are unique.
- nums is an ascending array that is possibly rotated.
- -10^4 <= target <= 10^4
class Solution {
public:
int search(vector nums, int target) {
int lo = 0, hi = nums.size() - 1;
while (lo <= hi) {
int mi = (lo + hi) >> 1;
if (nums[mi] == target) return mi;
if (nums[lo] <= nums[mi]) {
nums[lo] <= target && target < nums[mi] ? hi = mi - 1 : lo = mi + 1;
} else {
nums[mi] < target && target <= nums[hi] ? lo = mi + 1 : hi = mi - 1;
}
}
return -1;
}
};
