From f0b771af5c8d1fe5d5a298f00a2dbc05a1e23b42 Mon Sep 17 00:00:00 2001 From: Steve2020 <841532108@qq.com> Date: Wed, 30 Mar 2022 16:40:18 +0800 Subject: [PATCH 1/3] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=EF=BC=880131.=E5=88=86?= =?UTF-8?q?=E5=89=B2=E5=9B=9E=E6=96=87=E4=B8=B2.md=EF=BC=89=EF=BC=9A?= =?UTF-8?q?=E5=A2=9E=E5=8A=A0typescript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0131.分割回文串.md | 32 ++++++++++++++++++++++++++++++++ 1 file changed, 32 insertions(+) diff --git a/problems/0131.分割回文串.md b/problems/0131.分割回文串.md index f50f1c1d..10b747cb 100644 --- a/problems/0131.分割回文串.md +++ b/problems/0131.分割回文串.md @@ -450,6 +450,38 @@ var partition = function(s) { }; ``` +## TypeScript + +```typescript +function partition(s: string): string[][] { + function isPalindromeStr(s: string, left: number, right: number): boolean { + while (left < right) { + if (s[left++] !== s[right--]) { + return false; + } + } + return true; + } + function backTracking(s: string, startIndex: number, route: string[]): void { + let length: number = s.length; + if (length === startIndex) { + resArr.push(route.slice()); + return; + } + for (let i = startIndex; i < length; i++) { + if (isPalindromeStr(s, startIndex, i)) { + route.push(s.slice(startIndex, i + 1)); + backTracking(s, i + 1, route); + route.pop(); + } + } + } + const resArr: string[][] = []; + backTracking(s, 0, []); + return resArr; +}; +``` + ## C ```c From 7ee6bdbbfac43194c2ce3d5feb1f322824c3c7c7 Mon Sep 17 00:00:00 2001 From: h-yx-blog <2041290842@qq.com> Date: Thu, 31 Mar 2022 08:59:58 +0800 Subject: [PATCH 2/3] =?UTF-8?q?416=E5=88=86=E5=89=B2=E7=AD=89=E5=92=8C?= =?UTF-8?q?=E5=AD=90=E9=9B=86java=E7=89=88=E5=8F=A6=E4=B8=80=E7=A7=8D?= =?UTF-8?q?=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0416.分割等和子集.md | 69 +++++++++++++++++++++++++++++ 1 file changed, 69 insertions(+) diff --git a/problems/0416.分割等和子集.md b/problems/0416.分割等和子集.md index c8d9bc04..b24fb365 100644 --- a/problems/0416.分割等和子集.md +++ b/problems/0416.分割等和子集.md @@ -208,6 +208,75 @@ class Solution { } ``` +```java +public class Solution { + public static void main(String[] args) { + int num[] = {1,5,11,5}; + canPartition(num); + + } + public static boolean canPartition(int[] nums) { + int len = nums.length; + // 题目已经说非空数组，可以不做非空判断 + int sum = 0; + for (int num : nums) { + sum += num; + } + // 特判：如果是奇数，就不符合要求 + if ((sum %2 ) != 0) { + return false; + } + + int target = sum / 2; //目标背包容量 + // 创建二维状态数组，行：物品索引，列：容量（包括 0） + /* + dp[i][j]表示从数组的 [0, i] 这个子区间内挑选一些正整数 + 每个数只能用一次，使得这些数的和恰好等于 j。 + */ + boolean[][] dp = new boolean[len][target + 1]; + + // 先填表格第 0 行，第 1 个数只能让容积为它自己的背包恰好装满 （这里的dp[][]数组的含义就是“恰好”，所以就算容积比它大的也不要） + if (nums[0] <= target) { + dp[0][nums[0]] = true; + } + // 再填表格后面几行 + //外层遍历物品 + for (int i = 1; i < len; i++) { + //内层遍历背包 + for (int j = 0; j <= target; j++) { + // 直接从上一行先把结果抄下来，然后再修正 + dp[i][j] = dp[i - 1][j]; + + //如果某个物品单独的重量恰好就等于背包的重量，那么也是满足dp数组的定义的 + if (nums[i] == j) { + dp[i][j] = true; + continue; + } + //如果某个物品的重量小于j，那就可以看该物品是否放入背包 + //dp[i - 1][j]表示该物品不放入背包，如果在 [0, i - 1] 这个子区间内已经有一部分元素，使得它们的和为 j ，那么 dp[i][j] = true； + //dp[i - 1][j - nums[i]]表示该物品放入背包。如果在 [0, i - 1] 这个子区间内就得找到一部分元素，使得它们的和为 j - nums[i]。 + if (nums[i] < j) { + dp[i][j] = dp[i - 1][j] || dp[i - 1][j - nums[i]]; + } + } + } + for (int i = 0; i < len; i++) { + for (int j = 0; j <= target; j++) { + System.out.print(dp[i][j]+" "); + } + System.out.println(); + } + return dp[len - 1][target]; + } +} +//dp数组的打印结果 +false true false false false false false false false false false false +false true false false false true true false false false false false +false true false false false true true false false false false true +false true false false false true true false false false true true +``` + + 二维数组版本（易于理解）： ```Java class Solution { From 00dc57faaca1a77fd7c371f2975df5939a7edb0c Mon Sep 17 00:00:00 2001 From: MoonLight-Sherry <3397446353@qq.com> Date: Thu, 31 Mar 2022 10:34:34 +0800 Subject: [PATCH 3/3] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=E4=BA=860509.=E6=96=90?= =?UTF-8?q?=E6=B3=A2=E9=82=A3=E5=A5=91=E6=95=B0=E5=88=97Rust=E4=B8=A4?= =?UTF-8?q?=E7=A7=8D=E8=A7=A3=E6=B3=95=E7=9A=84=E4=BB=A3=E7=A0=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0509.斐波那契数.md | 25 ++++++++++++++++++++++++- 1 file changed, 24 insertions(+), 1 deletion(-) diff --git a/problems/0509.斐波那契数.md b/problems/0509.斐波那契数.md index c6ce76c0..e60b8a9e 100644 --- a/problems/0509.斐波那契数.md +++ b/problems/0509.斐波那契数.md @@ -277,7 +277,30 @@ int fib(int n){ return fib(n-1) + fib(n-2); } ``` +### Rust +动态规划: +```Rust +pub fn fib(n: i32) -> i32 { + let n = n as usize; + let mut dp = vec![0; 31]; + dp[1] = 1; + for i in 2..=n { + dp[i] = dp[i - 1] + dp[i - 2]; + } + dp[n] +} +``` - +递归实现： +```Rust +pub fn fib(n: i32) -> i32 { + //若n小于等于1，返回n + f n <= 1 { + return n; + } + //否则返回fib(n-1) + fib(n-2) + return fib(n - 1) + fib(n - 2); +} +``` -----------------------