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Commit 06533ad

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feat: add solutions to lc problem: No.1317 (doocs#2844)
1 parent 818daa9 commit 06533ad

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-11
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7 files changed

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-11
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‎solution/1300-1399/1317.Convert Integer to the Sum of Two No-Zero Integers/README.md‎

Lines changed: 41 additions & 3 deletions
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@@ -80,9 +80,9 @@ tags:
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### 方法一:直接枚举
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从 1ドル$ 开始枚举 $a,ドル判断 $a$ 和 $n - a$ 是否满足条件,如果满足则返回。
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从 1ドル$ 开始枚举 $a,ドル那么 $b = n - a$。对于每个 $a$ 和 $b,ドル我们将它们转换为字符串并且连接起来,然后判断是否包含字符 `'0'`,如果不包含,那么就找到了答案,返回 $[a, b]$。
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时间复杂度 $O(n\times \log n),ドル空间复杂度 $O(1)$。其中 $n$ 为题目给定的整数。
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时间复杂度 $O(n\times \log n),ドル其中 $n$ 为题目给定的整数。空间复杂度 $O(\log n)$。
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<!-- tabs:start -->
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@@ -141,13 +141,30 @@ func getNoZeroIntegers(n int) []int {
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}
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```
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#### TypeScript
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```ts
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function getNoZeroIntegers(n: number): number[] {
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for (let a = 1; ; ++a) {
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const b = n - a;
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if (!`${a}${b}`.includes('0')) {
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return [a, b];
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}
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- solution:start -->
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### 方法二
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### 方法二:直接枚举(另一种写法)
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在方法一中,我们将 $a$ 和 $b$ 转换为字符串并且连接起来,然后判断是否包含字符 `'0'`。这里我们可以通过一个函数 $f(x)$ 来判断 $x$ 是否包含字符 `'0'`,然后直接枚举 $a,ドル判断 $a$ 和 $b = n - a$ 是否都不包含字符 `'0'`,如果是,则找到了答案,返回 $[a, b]$。
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时间复杂度 $O(n \times \log n),ドル其中 $n$ 为题目给定的整数。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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}
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```
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#### TypeScript
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```ts
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function getNoZeroIntegers(n: number): number[] {
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const f = (x: number): boolean => {
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for (; x; x = (x / 10) | 0) {
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if (x % 10 === 0) {
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return false;
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}
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}
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return true;
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};
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for (let a = 1; ; ++a) {
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const b = n - a;
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if (f(a) && f(b)) {
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return [a, b];
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}
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->

‎solution/1300-1399/1317.Convert Integer to the Sum of Two No-Zero Integers/README_EN.md‎

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Original file line numberDiff line numberDiff line change
@@ -62,7 +62,11 @@ Note that there are other valid answers as [8, 3] that can be accepted.
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Direct Enumeration
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Starting from 1ドル,ドル we enumerate $a,ドル then $b = n - a$. For each $a$ and $b,ドル we convert them to strings and concatenate them, then check if they contain the character '0'. If they do not contain '0', we have found the answer and return $[a, b]$.
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The time complexity is $O(n \times \log n),ドル where $n$ is the integer given in the problem. The space complexity is $O(\log n)$.
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<!-- tabs:start -->
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}
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```
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#### TypeScript
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```ts
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function getNoZeroIntegers(n: number): number[] {
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for (let a = 1; ; ++a) {
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const b = n - a;
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if (!`${a}${b}`.includes('0')) {
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return [a, b];
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}
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- solution:start -->
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### Solution 2
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### Solution 2: Direct Enumeration (Alternative Approach)
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In Solution 1, we converted $a$ and $b$ into strings and concatenated them, then checked if they contained the character '0'. Here, we can use a function $f(x)$ to check whether $x$ contains the character '0', and then directly enumerate $a,ドル checking whether both $a$ and $b = n - a$ do not contain the character '0'. If they do not, we have found the answer and return $[a, b]$.
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The time complexity is $O(n \times \log n),ドル where $n$ is the integer given in the problem. The space complexity is $O(1)$.
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<!-- tabs:start -->
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@@ -218,6 +239,27 @@ func getNoZeroIntegers(n int) []int {
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}
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```
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#### TypeScript
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```ts
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function getNoZeroIntegers(n: number): number[] {
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const f = (x: number): boolean => {
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for (; x; x = (x / 10) | 0) {
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if (x % 10 === 0) {
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return false;
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}
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}
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return true;
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};
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for (let a = 1; ; ++a) {
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const b = n - a;
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if (f(a) && f(b)) {
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return [a, b];
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}
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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Original file line numberDiff line numberDiff line change
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function getNoZeroIntegers(n: number): number[] {
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for (let a = 1; ; ++a) {
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const b = n - a;
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if (!`${a}${b}`.includes('0')) {
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return [a, b];
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}
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}
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}
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function getNoZeroIntegers(n: number): number[] {
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const f = (x: number): boolean => {
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for (; x; x = (x / 10) | 0) {
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if (x % 10 === 0) {
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return false;
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}
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}
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return true;
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};
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for (let a = 1; ; ++a) {
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const b = n - a;
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if (f(a) && f(b)) {
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return [a, b];
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}
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}
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}

‎solution/3000-3099/3068.Find the Maximum Sum of Node Values/README.md‎

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```cpp
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class Solution {
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public:
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long long maximumValueSum(vector<int>& nums, int k,
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vector<vector<int>>& edges) {
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long long maximumValueSum(vector<int>& nums, int k, vector<vector<int>>& edges) {
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long long totalSum = 0;
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int count = 0;
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int positiveMin = INT_MAX;

‎solution/3000-3099/3068.Find the Maximum Sum of Node Values/README_EN.md‎

Lines changed: 1 addition & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -111,8 +111,7 @@ It can be shown that 9 is the maximum achievable sum of values.
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```cpp
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class Solution {
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public:
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long long maximumValueSum(vector<int>& nums, int k,
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vector<vector<int>>& edges) {
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long long maximumValueSum(vector<int>& nums, int k, vector<vector<int>>& edges) {
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long long totalSum = 0;
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int count = 0;
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int positiveMin = INT_MAX;

‎solution/3000-3099/3068.Find the Maximum Sum of Node Values/Solution.cpp‎

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class Solution {
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public:
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long long maximumValueSum(vector<int>& nums, int k,
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vector<vector<int>>& edges) {
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long long maximumValueSum(vector<int>& nums, int k, vector<vector<int>>& edges) {
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long long totalSum = 0;
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int count = 0;
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int positiveMin = INT_MAX;

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