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github.com | ||
golang.org | ||
gopkg.in | ||
tingle |
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package main | ||
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import "fmt" | ||
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// 树:节点node、根节点root、子节点child、叶节点leaf | ||
// 二叉树:每个节点最多有两个节点 | ||
// 完全二叉树 | ||
// 满二叉树 | ||
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// bigTopHeap 大顶堆 | ||
func bigTopHeap(input []int) []int { | ||
aroundFind := false | ||
lenInput := len(input) | ||
i := 0 | ||
for { | ||
if 2*(i+1) > lenInput { | ||
if !aroundFind { | ||
return input | ||
} | ||
i = 0 | ||
aroundFind = false | ||
continue | ||
} | ||
if input[i] < input[2*i+1] { | ||
tmp := input[i] | ||
input[i] = input[2*i+1] | ||
input[2*i+1] = tmp | ||
aroundFind = true | ||
} | ||
if 2*(i+1) < lenInput && input[i] < input[2*i+2] { | ||
tmp := input[i] | ||
input[i] = input[2*i+2] | ||
input[2*i+2] = tmp | ||
aroundFind = true | ||
} | ||
i++ | ||
} | ||
return input | ||
} | ||
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// lessTopHeap 小顶堆 | ||
func lessTopHeap(input []int) []int { | ||
aroundFind := false | ||
lenInput := len(input) | ||
i := 0 | ||
for { | ||
if 2*(i+1) > lenInput { | ||
if !aroundFind { | ||
return input | ||
} | ||
i = 0 | ||
aroundFind = false | ||
continue | ||
} | ||
if input[i] > input[2*i+1] { | ||
tmp := input[i] | ||
input[i] = input[2*i+1] | ||
input[2*i+1] = tmp | ||
aroundFind = true | ||
} | ||
if 2*(i+1) < lenInput && input[i] > input[2*i+2] { | ||
tmp := input[i] | ||
input[i] = input[2*i+2] | ||
input[2*i+2] = tmp | ||
aroundFind = true | ||
} | ||
i++ | ||
} | ||
return input | ||
} | ||
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func main() { | ||
input := []int{2, 7, 26, 25, 19, 17, 1, 90, 3, 36} | ||
fmt.Println(bigTopHeap(input)) | ||
fmt.Println(lessTopHeap(input)) | ||
} |
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package main | ||
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func main() { | ||
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} | ||
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func uniqueOccurrences(arr []int) bool { | ||
resMap := map[int]int{} | ||
for _, item := range arr { | ||
if _, ok := resMap[item]; ok { | ||
resMap[item]++ | ||
continue | ||
} | ||
resMap[item] = 1 | ||
} | ||
resNewMap := map[int]int{} | ||
for _, item := range resMap { | ||
if _, ok := resNewMap[item]; ok { | ||
return false | ||
} | ||
resNewMap[item] = 1 | ||
} | ||
return true | ||
} |
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package main | ||
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import "fmt" | ||
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type TreeNode struct { | ||
Val int | ||
Left *TreeNode | ||
Right *TreeNode | ||
} | ||
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func main() { | ||
two := &TreeNode{Val: 3} | ||
one := &TreeNode{Val: 2, Left: two} | ||
root := &TreeNode{Val: 1, Left: one} | ||
fmt.Println(preorderTraversal(root)) | ||
} | ||
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// type Stack struct { | ||
// data [] | ||
// } | ||
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/** | ||
* Definition for a binary tree node. | ||
* type TreeNode struct { | ||
* Val int | ||
* Left *TreeNode | ||
* Right *TreeNode | ||
* } | ||
*/ | ||
func preorderTraversal(root *TreeNode) []int { | ||
res := []int{} | ||
if root == nil { | ||
return res | ||
} | ||
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// 初始化栈 | ||
stack := []*TreeNode{} | ||
// 入栈 | ||
pushStack := func(stack []*TreeNode, val *TreeNode) []*TreeNode { | ||
return append(stack, val) | ||
} | ||
// 出栈 | ||
popStack := func(stack []*TreeNode) ([]*TreeNode, *TreeNode) { | ||
len := len(stack) | ||
if len == 0 { | ||
return []*TreeNode{}, nil | ||
} | ||
val := stack[len-1] | ||
return append(stack[:0], stack[:len-1]...), val | ||
} | ||
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// 根节点入栈 | ||
stack = pushStack(stack, root) | ||
node := &TreeNode{} | ||
for { | ||
if len(stack) == 0 { | ||
return res | ||
} | ||
stack, node = popStack(stack) | ||
res = append(res, node.Val) | ||
if node.Right != nil { | ||
stack = pushStack(stack, node.Right) | ||
} | ||
if node.Left != nil { | ||
stack = pushStack(stack, node.Left) | ||
} | ||
} | ||
} | ||
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/** | ||
* Definition for a binary tree node. | ||
* type TreeNode struct { | ||
* Val int | ||
* Left *TreeNode | ||
* Right *TreeNode | ||
* } | ||
*/ | ||
func preorderTraversalRecursion(root *TreeNode) []int { | ||
if root == nil { | ||
return []int{} | ||
} | ||
res := []int{root.Val} | ||
res = append(res, preorderTraversal(root.Left)...) | ||
res = append(res, preorderTraversal(root.Right)...) | ||
return res | ||
} |
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package leetcode | ||
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// ================ | ||
// leetcode题解 | ||
// 347. 前 K 个高频元素 | ||
// ================ | ||
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// 给定一个非空的整数数组,返回其中出现频率前 k 高的元素。 | ||
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// 示例 1: | ||
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// 输入: nums = [1,1,1,2,2,3], k = 2 | ||
// 输出: [1,2] | ||
// 示例 2: | ||
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// 输入: nums = [1], k = 1 | ||
// 输出: [1] | ||
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// 提示: | ||
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// 你可以假设给定的 k 总是合理的,且 1 ≤ k ≤ 数组中不相同的元素的个数。 | ||
// 你的算法的时间复杂度必须优于 O(n log n) , n 是数组的大小。 | ||
// 题目数据保证答案唯一,换句话说,数组中前 k 个高频元素的集合是唯一的。 | ||
// 你可以按任意顺序返回答案。 | ||
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// 来源:力扣(LeetCode) | ||
// 链接:https://leetcode-cn.com/problems/top-k-frequent-elements | ||
// 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。 | ||
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// 解法一:快排 | ||
func topKFrequent(nums []int, k int) []int { | ||
res := []int{} | ||
if len(nums) == 0 || k == 0 { | ||
return res | ||
} | ||
numsMap := map[int]int{} | ||
for _, v := range nums { | ||
if _, ok := numsMap[v]; ok { | ||
numsMap[v]++ | ||
continue | ||
} | ||
numsMap[v] = 1 | ||
} | ||
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resSlice := []int{} | ||
for _, v := range numsMap { | ||
resSlice = append(resSlice, v) | ||
} | ||
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// resSlice 快排 | ||
resSliceLen := len(resSlice) | ||
resSlice = quick(resSlice, 0, resSliceLen) | ||
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// 输出后k个(最大值) | ||
resSlice = append(resSlice[:0], resSlice[resSliceLen-k:]...) | ||
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resSliceMap := map[int]byte{} | ||
for _, v := range resSlice { | ||
resSliceMap[v] = '0' | ||
} | ||
for k, v := range numsMap { | ||
if _, ok := resSliceMap[v]; ok { | ||
res = append(res, k) | ||
} | ||
} | ||
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return res | ||
} | ||
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func quick(resSlice []int, left int, right int) (res []int) { | ||
if left >= right { | ||
return resSlice | ||
} | ||
base := left | ||
i := right | ||
j := left | ||
for { | ||
if i <= j { | ||
break | ||
} | ||
for i = right; i > base; i-- { | ||
if resSlice[i] < resSlice[base] { | ||
tmp := resSlice[i] | ||
resSlice[i] = resSlice[base] | ||
resSlice[base] = tmp | ||
base = i | ||
break | ||
} | ||
} | ||
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for j = left; j < base; j++ { | ||
if resSlice[j] > resSlice[base] { | ||
tmp := resSlice[j] | ||
resSlice[j] = resSlice[base] | ||
resSlice[base] = tmp | ||
base = j | ||
break | ||
} | ||
} | ||
} | ||
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resSlice = quick(resSlice, left, base-1) | ||
resSlice = quick(resSlice, base+1, right) | ||
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return resSlice | ||
} |
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// 输入一棵二叉树的根节点,判断该树是不是平衡二叉树。如果某二叉树中任意节点的左右子树的深度相差不超过1,那么它就是一棵平衡二叉树。 | ||
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// | ||
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// 示例 1: | ||
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// 给定二叉树 [3,9,20,null,null,15,7] | ||
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// 3 | ||
// / \ | ||
// 9 20 | ||
// / \ | ||
// 15 7 | ||
// 返回 true 。 | ||
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// 示例 2: | ||
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// 给定二叉树 [1,2,2,3,3,null,null,4,4] | ||
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// 1 | ||
// / \ | ||
// 2 2 | ||
// / \ | ||
// 3 3 | ||
// / \ | ||
// 4 4 | ||
// 返回 false 。 | ||
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// | ||
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// 限制: | ||
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// 1 <= 树的结点个数 <= 10000 | ||
// 注意:本题与主站 110 题相同:https://leetcode-cn.com/problems/balanced-binary-tree/ | ||
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// 来源:力扣(LeetCode) | ||
// 链接:https://leetcode-cn.com/problems/ping-heng-er-cha-shu-lcof | ||
// 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。 | ||
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type TreeNode struct { | ||
Val int | ||
Left *TreeNode | ||
Right *TreeNode | ||
} | ||
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/** | ||
* Definition for a binary tree node. | ||
* type TreeNode struct { | ||
* Val int | ||
* Left *TreeNode | ||
* Right *TreeNode | ||
* } | ||
*/ | ||
func isBalanced(root *TreeNode) bool { | ||
if root == nil { | ||
return false | ||
} | ||
if root.Left != nil { | ||
deep | ||
} | ||
} |
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