From 2fe9a1beeef252f8149978477e893c36833daf23 Mon Sep 17 00:00:00 2001 From: yanglbme Date: Fri, 31 Jan 2025 22:01:16 +0800 Subject: [PATCH] feat: add solutions to lc problems: No.2119,2177 --- .../README.md | 36 ++++++++++++++++++- .../README_EN.md | 36 ++++++++++++++++++- .../Solution.js | 7 ++++ .../Solution.rs | 5 +++ .../Solution.ts | 3 ++ .../README.md | 32 ++++++++++++++++- .../README_EN.md | 32 ++++++++++++++++- .../Solution.js | 11 ++++++ .../Solution.rs | 9 +++++ .../README.md | 2 +- 10 files changed, 168 insertions(+), 5 deletions(-) create mode 100644 solution/2100-2199/2119.A Number After a Double Reversal/Solution.js create mode 100644 solution/2100-2199/2119.A Number After a Double Reversal/Solution.rs create mode 100644 solution/2100-2199/2119.A Number After a Double Reversal/Solution.ts create mode 100644 solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/Solution.js create mode 100644 solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/Solution.rs diff --git a/solution/2100-2199/2119.A Number After a Double Reversal/README.md b/solution/2100-2199/2119.A Number After a Double Reversal/README.md index 117583e9746f4..a462fb9e6f900 100644 --- a/solution/2100-2199/2119.A Number After a Double Reversal/README.md +++ b/solution/2100-2199/2119.A Number After a Double Reversal/README.md @@ -62,7 +62,11 @@ tags: -### 方法一 +### 方法一：数学 + +如果数字是 $0$，或者数字的个位不是 $0$，那么反转两次后的数字一定和原数字相等。 + +时间复杂度 $O(1)$，空间复杂度 $O(1)$。 @@ -103,6 +107,36 @@ func isSameAfterReversals(num int) bool { } ``` +#### TypeScript + +```ts +function isSameAfterReversals(num: number): boolean { + return num === 0 || num % 10 !== 0; +} +``` + +#### Rust + +```rust +impl Solution { + pub fn is_same_after_reversals(num: i32) -> bool { + num == 0 || num % 10 != 0 + } +} +``` + +#### JavaScript + +```js +/** + * @param {number} num + * @return {boolean} + */ +var isSameAfterReversals = function (num) { + return num === 0 || num % 10 !== 0; +}; +``` + diff --git a/solution/2100-2199/2119.A Number After a Double Reversal/README_EN.md b/solution/2100-2199/2119.A Number After a Double Reversal/README_EN.md index b8c12ddd1000e..41dc16b2accd3 100644 --- a/solution/2100-2199/2119.A Number After a Double Reversal/README_EN.md +++ b/solution/2100-2199/2119.A Number After a Double Reversal/README_EN.md @@ -64,7 +64,11 @@ tags: -### Solution 1 +### Solution 1: Mathematics + +If the number is $0$, or the last digit of the number is not $0$, then the number after reversing twice will be the same as the original number. + +The time complexity is $O(1)$, and the space complexity is $O(1)$. @@ -105,6 +109,36 @@ func isSameAfterReversals(num int) bool { } ``` +#### TypeScript + +```ts +function isSameAfterReversals(num: number): boolean { + return num === 0 || num % 10 !== 0; +} +``` + +#### Rust + +```rust +impl Solution { + pub fn is_same_after_reversals(num: i32) -> bool { + num == 0 || num % 10 != 0 + } +} +``` + +#### JavaScript + +```js +/** + * @param {number} num + * @return {boolean} + */ +var isSameAfterReversals = function (num) { + return num === 0 || num % 10 !== 0; +}; +``` + diff --git a/solution/2100-2199/2119.A Number After a Double Reversal/Solution.js b/solution/2100-2199/2119.A Number After a Double Reversal/Solution.js new file mode 100644 index 0000000000000..e971a7af9ec2c --- /dev/null +++ b/solution/2100-2199/2119.A Number After a Double Reversal/Solution.js @@ -0,0 +1,7 @@ +/** + * @param {number} num + * @return {boolean} + */ +var isSameAfterReversals = function (num) { + return num === 0 || num % 10 !== 0; +}; diff --git a/solution/2100-2199/2119.A Number After a Double Reversal/Solution.rs b/solution/2100-2199/2119.A Number After a Double Reversal/Solution.rs new file mode 100644 index 0000000000000..516887ac60f3f --- /dev/null +++ b/solution/2100-2199/2119.A Number After a Double Reversal/Solution.rs @@ -0,0 +1,5 @@ +impl Solution { + pub fn is_same_after_reversals(num: i32) -> bool { + num == 0 || num % 10 != 0 + } +} diff --git a/solution/2100-2199/2119.A Number After a Double Reversal/Solution.ts b/solution/2100-2199/2119.A Number After a Double Reversal/Solution.ts new file mode 100644 index 0000000000000..6c0cd001ad95e --- /dev/null +++ b/solution/2100-2199/2119.A Number After a Double Reversal/Solution.ts @@ -0,0 +1,3 @@ +function isSameAfterReversals(num: number): boolean { + return num === 0 || num % 10 !== 0; +} diff --git a/solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/README.md b/solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/README.md index 69bf1511aff3b..b5ac84cbc3087 100644 --- a/solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/README.md +++ b/solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/README.md @@ -54,7 +54,7 @@ tags: ### 方法一：数学 -假设三个连续的整数分别为 $x-1$, $x$, $x+1$，则它们的和为 $3x$，因此 $num$ 必须是 $3$ 的倍数。如果 $num$ 不是 $3$ 的倍数，则无法表示成三个连续整数的和，返回空数组。否则，令 $x = \frac{num}{3}$，则 $x-1$, $x$, $x+1$ 就是三个连续整数，它们的和为 $num$。 +我们假设三个连续的整数分别为 $x-1$, $x$, $x+1$，则它们的和为 $3x$，因此 $\textit{num}$ 必须是 $3$ 的倍数。如果 $\textit{num}$ 不是 $3$ 的倍数，则无法表示成三个连续整数的和，返回空数组。否则，令 $x = \frac{\textit{num}}{3}$，则 $x-1$, $x$, $x+1$ 就是三个连续整数，它们的和为 $\textit{num}$。 时间复杂度 $O(1)$，空间复杂度 $O(1)$。 @@ -122,6 +122,36 @@ function sumOfThree(num: number): number[] { } ``` +#### Rust + +```rust +impl Solution { + pub fn sum_of_three(num: i64) -> Vec { + if num % 3 != 0 { + return Vec::new(); + } + let x = num / 3; + vec![x - 1, x, x + 1] + } +} +``` + +#### JavaScript + +```js +/** + * @param {number} num + * @return {number[]} + */ +var sumOfThree = function (num) { + if (num % 3) { + return []; + } + const x = Math.floor(num / 3); + return [x - 1, x, x + 1]; +}; +``` + diff --git a/solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/README_EN.md b/solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/README_EN.md index e496c2413f7c7..c9c665e1eba37 100644 --- a/solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/README_EN.md +++ b/solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/README_EN.md @@ -54,7 +54,7 @@ tags: ### Solution 1: Mathematics -Assume the three consecutive integers are $x-1$, $x$, and $x+1$. Their sum is $3x$, so $num$ must be a multiple of $3$. If $num$ is not a multiple of $3$, it cannot be expressed as the sum of three consecutive integers, and we return an empty array. Otherwise, let $x = \frac{num}{3}$, then $x-1$, $x$, and $x+1$ are the three consecutive integers whose sum is $num$. +Assume the three consecutive integers are $x-1$, $x$, and $x+1$. Their sum is $3x$, so $\textit{num}$ must be a multiple of $3$. If $\textit{num}$ is not a multiple of $3$, it cannot be represented as the sum of three consecutive integers, and we return an empty array. Otherwise, let $x = \frac{\textit{num}}{3}$, then $x-1$, $x$, and $x+1$ are the three consecutive integers whose sum is $\textit{num}$. The time complexity is $O(1)$, and the space complexity is $O(1)$. @@ -122,6 +122,36 @@ function sumOfThree(num: number): number[] { } ``` +#### Rust + +```rust +impl Solution { + pub fn sum_of_three(num: i64) -> Vec { + if num % 3 != 0 { + return Vec::new(); + } + let x = num / 3; + vec![x - 1, x, x + 1] + } +} +``` + +#### JavaScript + +```js +/** + * @param {number} num + * @return {number[]} + */ +var sumOfThree = function (num) { + if (num % 3) { + return []; + } + const x = Math.floor(num / 3); + return [x - 1, x, x + 1]; +}; +``` + diff --git a/solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/Solution.js b/solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/Solution.js new file mode 100644 index 0000000000000..548dbf71cc417 --- /dev/null +++ b/solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/Solution.js @@ -0,0 +1,11 @@ +/** + * @param {number} num + * @return {number[]} + */ +var sumOfThree = function (num) { + if (num % 3) { + return []; + } + const x = Math.floor(num / 3); + return [x - 1, x, x + 1]; +}; diff --git a/solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/Solution.rs b/solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/Solution.rs new file mode 100644 index 0000000000000..1079c374a223a --- /dev/null +++ b/solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/Solution.rs @@ -0,0 +1,9 @@ +impl Solution { + pub fn sum_of_three(num: i64) -> Vec { + if num % 3 != 0 { + return Vec::new(); + } + let x = num / 3; + vec![x - 1, x, x + 1] + } +} diff --git a/solution/2600-2699/2674.Split a Circular Linked List/README.md b/solution/2600-2699/2674.Split a Circular Linked List/README.md index 0abbed8f317f7..80957a399b316 100644 --- a/solution/2600-2699/2674.Split a Circular Linked List/README.md +++ b/solution/2600-2699/2674.Split a Circular Linked List/README.md @@ -46,7 +46,7 @@ tags:

提示：

    -
  • list 中的节点数范围为 [2, 105]
    • +
  • list 中的节点数范围为 [2, 105]
  • 0 <= Node.val <= 109
  • LastNode.next = FirstNode ，其中 LastNode 是链表的最后一个节点，FirstNode 是第一个节点。