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commontypes.go
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/*
Common type definitions, such as TreeNode, for LeetCode solutions
*/
package leetcode
/**
* Definition for singly-linked list.
*/
type ListNode struct {
Val int
Next *ListNode
}
/**
* Definition for a binary tree node.
*/
type TreeNode struct {
Val int
Left *TreeNode
Right *TreeNode
}
// This is the interface that allows for creating nested lists.
// You should not implement it, or speculate about its implementation
type NestedInteger struct {
}
// Return true if this NestedInteger holds a single integer, rather than a nested list.
func (n NestedInteger) IsInteger() bool { return true }
// Return the single integer that this NestedInteger holds, if it holds a single integer
// The result is undefined if this NestedInteger holds a nested list
// So before calling this method, you should have a check
func (n NestedInteger) GetInteger() int { return 0 }
// Set this NestedInteger to hold a single integer.
func (n *NestedInteger) SetInteger(value int) {}
// Set this NestedInteger to hold a nested list and adds a nested integer to it.
func (n *NestedInteger) Add(elem NestedInteger) {}
// Return the nested list that this NestedInteger holds, if it holds a nested list
// The list length is zero if this NestedInteger holds a single integer
// You can access NestedInteger's List element directly if you want to modify it
func (n NestedInteger) GetList() []*NestedInteger { return nil }
type trieNode struct {
children [26]*trieNode
isWord bool
}
// Interval is the definition for an Interval.
type Interval struct {
Start int
End int
}
// TreeNodeWithNext is renamed the tree node type for LeetCode 116/117
type TreeNodeWithNext struct {
Val int
Left *TreeNodeWithNext
Right *TreeNodeWithNext
Next *TreeNodeWithNext
}
type MinHeap []int
func (h MinHeap) Len() int { return len(h) }
func (h MinHeap) Less(i, j int) bool { return h[i] < h[j] }
func (h MinHeap) Swap(i, j int) { h[i], h[j] = h[j], h[i] }
func (h *MinHeap) Push(x any) {
*h = append(*h, x.(int))
}
func (h *MinHeap) Pop() any {
old := *h
n := len(old)
x := old[n-1]
*h = old[0 : n-1]
return x
}