规范格式

leetcode/0239.Sliding-Window-Maximum/239. Sliding Window Maximum.go

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+package leetcode
+
+// 解法一 暴力解法 O(nk)
+func maxSlidingWindow1(a []int, k int) []int {
+	res := make([]int, 0, k)
+	n := len(a)
+	if n == 0 {
+		return []int{}
+	}
+	for i := 0; i <= n-k; i++ {
+		max := a[i]
+		for j := 1; j < k; j++ {
+			if max < a[i+j] {
+				max = a[i+j]
+			}
+		}
+		res = append(res, max)
+	}
+
+	return res
+}
+
+// 解法二 双端队列 Deque
+func maxSlidingWindow(nums []int, k int) []int {
+	if len(nums) == 0 || len(nums) < k {
+		return make([]int, 0)
+	}
+	window := make([]int, 0, k) // store the index of nums
+	result := make([]int, 0, len(nums)-k+1)
+	for i, v := range nums { // if the left-most index is out of window, remove it
+		if i >= k && window[0] <= i-k {
+			window = window[1:len(window)]
+		}
+		for len(window) > 0 && nums[window[len(window)-1]] < v { // maintain window
+			window = window[0 : len(window)-1]
+		}
+		window = append(window, i) // store the index of nums
+		if i >= k-1 {
+			result = append(result, nums[window[0]]) // the left-most is the index of max value in nums
+		}
+	}
+	return result
+}

leetcode/0239.Sliding-Window-Maximum/239. Sliding Window Maximum_test.go

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+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question239 struct {
+	para239
+	ans239
+}
+
+// para 是参数
+// one 代表第一个参数
+type para239 struct {
+	one []int
+	k   int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans239 struct {
+	one []int
+}
+
+func Test_Problem239(t *testing.T) {
+
+	qs := []question239{
+
+		question239{
+			para239{[]int{1, 3, -1, -3, 5, 3, 6, 7}, 3},
+			ans239{[]int{3, 3, 5, 5, 6, 7}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 239------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans239, q.para239
+		fmt.Printf("【input】:%v       【output】:%v\n", p, maxSlidingWindow(p.one, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0239.Sliding-Window-Maximum/README.md

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+# [239. Sliding Window Maximum](https://leetcode.com/problems/sliding-window-maximum/)
+
+
+
+## 题目
+
+Given an array *nums*, there is a sliding window of size *k* which is moving from the very left of the array to the very right. You can only see the *k* numbers in the window. Each time the sliding window moves right by one position. Return the max sliding window.
+
+**Example:**
+
+    Input: nums = [1,3,-1,-3,5,3,6,7], and k = 3
+    Output: [3,3,5,5,6,7] 
+    Explanation: 
+    
+    Window position                Max
+    ---------------               -----
+    [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]      7
+
+**Note:**
+
+You may assume *k* is always valid, 1 ≤ k ≤ input array's size for non-empty array.
+
+**Follow up:**
+
+Could you solve it in linear time?
+
+
+## 题目大意
+
+给定一个数组 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 k 内的数字。滑动窗口每次只向右移动一位。返回滑动窗口最大值。
+
+
+## 解题思路
+
+- 给定一个数组和一个窗口为 K 的窗口，当窗口从数组的左边滑动到数组右边的时候，输出每次移动窗口以后，在窗口内的最大值。
+- 这道题最暴力的方法就是 2 层循环，时间复杂度 O(n * K)。
+- 另一种思路是用优先队列，每次窗口以后的时候都向优先队列里面新增一个节点，并删除一个节点。时间复杂度是 O(n * log n)
+- 最优的解法是用双端队列，队列的一边永远都存的是窗口的最大值，队列的另外一个边存的是比最大值小的值。队列中最大值左边的所有值都出队。在保证了双端队列的一边即是最大值以后，时间复杂度是 O(n)，空间复杂度是 O(K)
+