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Merged
merged 3 commits into from
Nov 22, 2023
Merged

feat: i18n update #39

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f74999f
feat: i18n update
rennzhang Nov 22, 2023
a3970cc
feat: i18n update
rennzhang Nov 22, 2023
00f3710
fix: fixed i18n for roadmap
rennzhang Nov 22, 2023
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24 changes: 18 additions & 6 deletions src/codeTemplates/backtrack.js
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Original file line number Diff line number Diff line change
Expand Up @@ -65,18 +65,30 @@ module.exports = () => ({
text: `
const visited = {}
function backtrack(i) {
if (满足特定条件）{
// 如果满足条件
if (Meet certain conditions) {
// 返回结果 or 退出搜索空间
// return result or exit search space
}

visited[i] = true // 将当前状态标为已搜索
dosomething(i) // 对i做一些操作
for (根据i能到达的下个状态j) {
if (!visited[j]) { // 如果状态j没有被搜索过
// 将当前状态标为已搜索
// mark the current state as searched
visited[i] = true
// 对i做一些操作
// do something with i
dosomething(i)

// for (根据i能到达的下个状态j) {
for (The next state j that can be reached based on i.) {
// 如果状态j没有被搜索过
// if state j has not been searched
if (!visited[j]) {
dfs(j)
}
}
undo(i) // 恢复i
// 恢复i
// restore i
undo(i)
}
backtrack(0)
`,
Expand Down
11 changes: 10 additions & 1 deletion src/codeTemplates/bfs.js
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Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -32,15 +32,20 @@ export default () => ({
class Solution:
def bfs(k):
# 使用双端队列，而不是数组。因为数组从头部删除元素的时间复杂度为 N，双端队列的底层实现其实是链表。
# Utilize a double-ended queue instead of an array, as the time complexity for removing elements from the head of an array is O(N), whereas a double-ended queue, implemented as a linked list, offers a more efficient alternative.
queue = collections.deque([root])
# 记录层数
# Record the level or depth.
steps = 0
# 需要返回的节点
# The nodes to return.
ans = []
# 队列不空，生命不止！
# While the queue is not empty, we continue.
while queue:
size = len(queue)
# 遍历当前层的所有节点
# Traverse all the nodes in the current level.
for _ in range(size):
node = queue.popleft()
if (steps == k) ans.append(node)
Expand All @@ -49,6 +54,7 @@ export default () => ({
if node.left:
queue.append(node.left)
# 遍历完当前层所有的节点后 steps + 1
# After traversing all the nodes in the current level, steps + 1.
steps += 1
return ans
`,
Expand All @@ -72,12 +78,15 @@ export default () => ({
class Solution:
def bfs(k):
# 使用双端队列，而不是数组。因为数组从头部删除元素的时间复杂度为 N，双端队列的底层实现其实是链表。
# Utilize a double-ended queue instead of an array, as the time complexity for removing elements from the head of an array is O(N), whereas a double-ended queue, implemented as a linked list, offers a more efficient alternative.
queue = collections.deque([root])
# 队列不空，生命不止！
# While the queue is not empty, we continue.
while queue:
node = queue.popleft()
# 由于没有记录 steps，因此我们肯定是不需要根据层的信息去判断的。否则就用带层的模板了。
if (node 是我们要找到的) return node
# Since we don't record steps, we don't need to judge based on the level information. Otherwise, we would use the template with level information.
if (node is what we are looking for) return node
if node.right:
queue.append(node.right)
if node.left:
Expand Down
50 changes: 48 additions & 2 deletions src/codeTemplates/binarySearch.js
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Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -20,6 +20,7 @@ module.exports = () => ({
text: `
public int binarySearch(int[] nums, int target) {
// 左右都闭合的区间 [l, r]
// A closed interval [l, r] with both ends inclusive.
int left = 0;
int right = nums.length - 1;

Expand All @@ -29,9 +30,11 @@ module.exports = () => ({
return mid;
if (nums[mid] < target)
// 搜索区间变为 [mid+1, right]
// Narrow down the search range to [mid+1, right]
left = mid + 1;
if (nums[mid] > target)
// 搜索区间变为 [left, mid - 1]
// Narrow down the search range to [left, mid - 1]
right = mid - 1;
}
return -1;
Expand All @@ -42,13 +45,16 @@ module.exports = () => ({
text: `
def binarySearch(nums, target):
# 左右都闭合的区间 [l, r]
# A closed interval [l, r] with both ends inclusive.
l, r = 0, len(nums) - 1
while l <= r:
mid = (left + right) >> 1
if nums[mid] == target: return mid
# 搜索区间变为 [mid+1, right]
# Narrow down the search range to [mid+1, right]
if nums[mid] < target: l = mid + 1
# 搜索区间变为 [left, mid - 1]
# Narrow down the search range to [left, mid - 1]
if nums[mid] > target: r = mid - 1
return -1`,
},
Expand All @@ -63,9 +69,11 @@ module.exports = () => ({
if (nums[mid] == target) return mid;
if (nums[mid] < target)
// 搜索区间变为 [mid+1, right]
// Narrow down the search range to [mid+1, right]
left = mid + 1;
if (nums[mid] > target)
// 搜索区间变为 [left, mid - 1]
// Narrow down the search range to [left, mid - 1]
right = mid - 1;
}
return -1;
Expand All @@ -83,9 +91,11 @@ module.exports = () => ({
int mid = left + ((right - left) >> 1);
if(nums[mid] == target){ return mid; }
// 搜索区间变为 [mid+1, right]
// Narrow down the search range to [mid+1, right]
else if(nums[mid] < target)
left = mid + 1;
// 搜索区间变为 [left, mid - 1]
// Narrow down the search range to [left, mid - 1]
else
right = mid - 1;
}
Expand Down Expand Up @@ -116,20 +126,24 @@ module.exports = () => ({
text: `
public int binarySearchLeft(int[] nums, int target) {
// 搜索区间为 [left, right]
// A closed interval [left, right] with both ends inclusive.
int left = 0;
int right = nums.length - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] < target) {
// 搜索区间变为 [mid+1, right]
// Narrow down the search range to [mid+1, right]
left = mid + 1;
}
if (nums[mid] >= target) {
// 搜索区间变为 [left, mid-1]
// Narrow down the search range to [left, mid-1]
right = mid - 1;
}
}
// 检查是否越界
// Check if it is out of bounds
if (left >= nums.length || nums[left] != target)
return -1;
return left;
Expand All @@ -140,12 +154,15 @@ module.exports = () => ({
text: `
def binarySearchLeft(nums, target):
# 左右都闭合的区间 [l, r]
# A closed interval [l, r] with both ends inclusive.
l, r = 0, len(nums) - 1
while l <= r:
mid = (l + r) >> 1
# 搜索区间变为 [mid+1, right]
# Narrow down the search range to [mid+1, right]
if nums[mid] < target: l = mid + 1
# 搜索区间变为 [left, mid - 1]
# Narrow down the search range to [left, mid - 1]
if nums[mid] >= target: r = mid - 1
if l >= len(nums) or nums[l] != target: return -1
return l`,
Expand All @@ -160,12 +177,15 @@ module.exports = () => ({
const mid = Math.floor(left + (right - left) / 2);
if (nums[mid] < target)
// 搜索区间变为 [mid+1, right]
// Narrow down the search range to [mid+1, right]
left = mid + 1;
if (nums[mid] >= target)
// 搜索区间变为 [left, mid - 1]
// Narrow down the search range to [left, mid - 1]
right = mid - 1;
}
// 检查是否越界
// Check if it is out of bounds
if (left >= nums.length || nums[left] != target) return -1;
return left;
}`,
Expand All @@ -175,23 +195,28 @@ module.exports = () => ({
text: `
int binarySearchLeft(vector<int>& nums, int target) {
// 搜索区间为 [left, right]
// The search interval is [left, right].
int left = 0, right = nums.size() - 1;
while (left <= right) {
int mid = left + ((right - left) >> 1);
if (nums[mid] == target) {
// 收缩右边界
// Narrow down the right boundary
right = mid - 1;
}
if (nums[mid] < target) {
// 搜索区间变为 [mid+1, right]
// Narrow down the search range to [mid+1, right]
left = mid + 1;
}
if (nums[mid] > target) {
// 搜索区间变为 [left, mid-1]
// Narrow down the search range to [left, mid-1]
right = mid - 1;
}
}
// 检查是否越界
// Check if it is out of bounds
if (left >= nums.size() || nums[left] != target)
return -1;
return left;
Expand Down Expand Up @@ -221,20 +246,24 @@ module.exports = () => ({
text: `
public int binarySearchRight(int[] nums, int target) {
// 搜索区间为 [left, right]
// A closed interval [left, right] with both ends inclusive.
int left = 0
int right = nums.length - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] <= target) {
// 搜索区间变为 [mid+1, right]
// Narrow down the search range to [mid+1, right]
left = mid + 1;
}
if (nums[mid] > target) {
// 搜索区间变为 [left, mid-1]
// Narrow down the search range to [left, mid-1]
right = mid - 1;
}
}
// 检查是否越界
// Check if it is out of bounds
if (right < 0 || nums[right] != target)
return -1;
return right;
Expand All @@ -245,12 +274,15 @@ module.exports = () => ({
text: `
def binarySearchRight(nums, target):
# 左右都闭合的区间 [l, r]
# A closed interval [l, r] with both ends inclusive.
l, r = 0, len(nums) - 1
while l <= r:
mid = (l + r) >> 1
# 搜索区间变为 [mid+1, right]
# Narrow down the search range to [mid+1, right]
if nums[mid] <= target: l = mid + 1
# 搜索区间变为 [left, mid - 1]
# Narrow down the search range to [left, mid - 1]
if nums[mid] > target: r = mid - 1
if r < 0 or nums[r] != target: return -1
return r`,
Expand All @@ -265,12 +297,15 @@ module.exports = () => ({
const mid = Math.floor(left + (right - left) / 2);
if (nums[mid] <= target)
// 搜索区间变为 [mid+1, right]
// Narrow down the search range to [mid+1, right]
left = mid + 1;
if (nums[mid] > target)
// 搜索区间变为 [left, mid - 1]
// Narrow down the search range to [left, mid - 1]
right = mid - 1;
}
// 检查是否越界
// Check if it is out of bounds
if (right < 0 || nums[right] != target) return -1;
return right;
}`,
Expand All @@ -280,23 +315,28 @@ module.exports = () => ({
text: `
int binarySearchRight(vector<int>& nums, int target) {
// 搜索区间为 [left, right]
// The search interval is [left, right].
int left = 0, right = nums.size() - 1;
while (left <= right) {
int mid = left + ((right - left) >> 1);
if (nums[mid] == target) {
// 收缩左边界
// Narrow down the left boundary
left = mid + 1;
}
if (nums[mid] < target) {
// 搜索区间变为 [mid+1, right]
// Narrow down the search range to [mid+1, right]
left = mid + 1;
}
if (nums[mid] > target) {
// 搜索区间变为 [left, mid-1]
// Narrow down the search range to [left, mid-1]
right = mid - 1;
}
}
// 检查是否越界
// Check if it is out of bounds
if (right < 0 || nums[right] != target)
return -1;
return right;
Expand All @@ -315,8 +355,10 @@ module.exports = () => ({
text: `
def bisect_left(nums, x):
# 内置 api
# built-in API
bisect.bisect_left(nums, x)
# 手写
# Manually write
l, r = 0, len(A) - 1
while l <= r:
mid = (l + r) // 2
Expand All @@ -330,13 +372,14 @@ module.exports = () => ({
text: `
/**
* @author suukii
* @description 寻找最左插入位置
* @description ${t("Locale.codeTemplate.binarySearch.item4")}
* @param {number[]} nums
* @param {number} x
* @returns {number}
*/
function searchInsertLeft(nums, x) {
// 题意转换一下，其实就是寻找第一个“大于等于” x 的数字，返回它的下标
// Change the meaning of the question, in fact, it is to find the first number "greater than or equal to" x and return its subscript
let left = 0;
let right = nums.length - 1;

Expand Down Expand Up @@ -369,8 +412,10 @@ function searchInsertLeft(nums, x) {
text: `
def bisect_right(nums, x):
# 内置 api
# built-in API
bisect.bisect_right(nums, x)
# 手写
# Manually write
l, r = 0, len(A) - 1
while l <= r:
mid = (l + r) // 2
Expand All @@ -383,13 +428,14 @@ function searchInsertLeft(nums, x) {
language: "JS",
text: `
/**@author suukii
* @description 寻找最右插入位置
* @description ${t("Locale.codeTemplate.binarySearch.item5")}
* @param {number[]} nums
* @param {number} x
* @returns {number}
*/
function searchInsertRight(nums, x) {
// 题意转换一下，其实就是寻找第一个“大于” x 的数字，返回它的下标
// Change the meaning of the question, in fact, it is to find the first number "greater than" x and return its subscript
let left = 0;
let right = nums.length - 1;

Expand Down
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