feat: add solutions to lc problem: No.1953

lcp/LCP 34. 二叉树染色/README.md

@@ -18,8 +18,7 @@ edit_url: https://github.com/doocs/leetcode/edit/main/lcp/LCP%2034.%20%E4%BA%8C%
 >
 > 输出：`12`
 >
-> 解释：`结点 5、3、4 染成蓝色，获得最大的价值 5+3+4=12`
-> ![image.png](https://fastly.jsdelivr.net/gh/doocs/leetcode@main/lcp/LCP%2034.%20%E4%BA%8C%E5%8F%89%E6%A0%91%E6%9F%93%E8%89%B2/images/1616126267-BqaCRj-image.png)
+> 解释：`结点 5、3、4 染成蓝色，获得最大的价值 5+3+4=12` > ![image.png](https://fastly.jsdelivr.net/gh/doocs/leetcode@main/lcp/LCP%2034.%20%E4%BA%8C%E5%8F%89%E6%A0%91%E6%9F%93%E8%89%B2/images/1616126267-BqaCRj-image.png)
 
 **示例 2：**
 

lcp/LCP 52. 二叉搜索树染色/README.md

@@ -38,8 +38,7 @@ edit_url: https://github.com/doocs/leetcode/edit/main/lcp/LCP%2052.%20%E4%BA%8C%
 
 **示例 2：**
 
-> 输入：`root = [4,2,7,1,null,5,null,null,null,null,6]`
-> `ops = [[0,2,2],[1,1,5],[0,4,5],[1,5,7]]`
+> 输入：`root = [4,2,7,1,null,5,null,null,null,null,6]` > `ops = [[0,2,2],[1,1,5],[0,4,5],[1,5,7]]`
 >
 > 输出：`5`
 >

lcp/LCP 61. 气温变化趋势/README.md

@@ -28,8 +28,7 @@ edit_url: https://github.com/doocs/leetcode/edit/main/lcp/LCP%2061.%20%E6%B0%94%
 >
 > 输出：`2`
 >
-> 解释：如下表所示， 第 `2～4` 天两地气温变化趋势相同，且持续时间最长，因此返回 `4-2=2`
-> ![image.png](https://fastly.jsdelivr.net/gh/doocs/leetcode@main/lcp/LCP%2061.%20%E6%B0%94%E6%B8%A9%E5%8F%98%E5%8C%96%E8%B6%8B%E5%8A%BF/images/1663902654-hlrSvs-image.png){:width=1000px}
+> 解释：如下表所示， 第 `2～4` 天两地气温变化趋势相同，且持续时间最长，因此返回 `4-2=2` > ![image.png](https://fastly.jsdelivr.net/gh/doocs/leetcode@main/lcp/LCP%2061.%20%E6%B0%94%E6%B8%A9%E5%8F%98%E5%8C%96%E8%B6%8B%E5%8A%BF/images/1663902654-hlrSvs-image.png){:width=1000px}
 
 **示例 2：**
 

lcp/LCP 74. 最强祝福力场/README.md

@@ -37,8 +37,7 @@ edit_url: https://github.com/doocs/leetcode/edit/main/lcp/LCP%2074.%20%E6%9C%80%
 > 输出：`3`
 >
 > 解释：如下图所示，
-> `forceField[0]、forceField[1]、forceField[3]` 重叠的区域力场强度最大，返回 `3`
-> ![image.png](https://fastly.jsdelivr.net/gh/doocs/leetcode@main/lcp/LCP%2074.%20%E6%9C%80%E5%BC%BA%E7%A5%9D%E7%A6%8F%E5%8A%9B%E5%9C%BA/images/1681805437-HQkyZS-image.png){:width=500px}
+> `forceField[0]、forceField[1]、forceField[3]` 重叠的区域力场强度最大，返回 `3` > ![image.png](https://fastly.jsdelivr.net/gh/doocs/leetcode@main/lcp/LCP%2074.%20%E6%9C%80%E5%BC%BA%E7%A5%9D%E7%A6%8F%E5%8A%9B%E5%9C%BA/images/1681805437-HQkyZS-image.png){:width=500px}
 
 **提示：**
 

lcp/LCP 80. 生物进化录/README.md

@@ -33,8 +33,7 @@ edit_url: https://github.com/doocs/leetcode/edit/main/lcp/LCP%2080.%20%E7%94%9F%
 > 解释：树结构如下图所示，共存在 2 种记录方案：
 > 第 1 种方案为：0(记录编号 1 的节点) -> 1(回退至节点 0) -> 0(记录编号 2 的节点) -> 0((记录编号 3 的节点))
 > 第 2 种方案为：0(记录编号 2 的节点) -> 0(记录编号 3 的节点) -> 1(回退至节点 2) -> 1(回退至节点 0) -> 0(记录编号 1 的节点)
-> 返回字典序更小的 `"00110"`
-> ![image.png](https://fastly.jsdelivr.net/gh/doocs/leetcode@main/lcp/LCP%2080.%20%E7%94%9F%E7%89%A9%E8%BF%9B%E5%8C%96%E5%BD%95/images/1682319485-cRVudI-image.png){:width=120px}![进化 (3).gif](<https://fastly.jsdelivr.net/gh/doocs/leetcode@main/lcp/LCP%2080.%20%E7%94%9F%E7%89%A9%E8%BF%9B%E5%8C%96%E5%BD%95/images/1682412701-waHdnm-%E8%BF%9B%E5%8C%96%20(3).gif>){:width=320px}
+> 返回字典序更小的 `"00110"` > ![image.png](https://fastly.jsdelivr.net/gh/doocs/leetcode@main/lcp/LCP%2080.%20%E7%94%9F%E7%89%A9%E8%BF%9B%E5%8C%96%E5%BD%95/images/1682319485-cRVudI-image.png){:width=120px}![进化 (3).gif](<https://fastly.jsdelivr.net/gh/doocs/leetcode@main/lcp/LCP%2080.%20%E7%94%9F%E7%89%A9%E8%BF%9B%E5%8C%96%E5%BD%95/images/1682412701-waHdnm-%E8%BF%9B%E5%8C%96%20(3).gif>){:width=320px}
 
 **示例 2：**
 

solution/1900-1999/1953.Maximum Number of Weeks for Which You Can Work/README.md

@@ -75,7 +75,15 @@ tags:
 
 ## 解法
 
-### 方法一
+### 方法一：贪心
+
+我们考虑什么情况下不能完成所有阶段任务。如果存在一个项目 $i$，它的阶段任务数大于其余所有项目的阶段任务数之和再加 $1$，那么就不能完成所有阶段任务。否则，我们一定可以通过不同项目之间来回穿插的方式完成所有阶段任务。
+
+我们记所有项目的阶段任务数之和为 $s$，最大的阶段任务数为 $mx$，那么其余所有项目的阶段任务数之和为 $rest = s - mx$。
+
+如果 $mx \gt rest + 1$，那么就不能完成所有阶段任务，最多只能完成 $rest \times 2 + 1$ 个阶段任务。否则，我们可以完成所有阶段任务，数量为 $s$。
+
+时间复杂度 $O(n)$，其中 $n$ 为项目数。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -129,6 +137,15 @@ func numberOfWeeks(milestones []int) int64 {
 }
 ```
 
+```ts
+function numberOfWeeks(milestones: number[]): number {
+    const mx = Math.max(...milestones);
+    const s = milestones.reduce((a, b) => a + b, 0);
+    const rest = s - mx;
+    return mx > rest + 1 ? rest * 2 + 1 : s;
+}
+```
+
 <!-- tabs:end -->
 
 <!-- end -->

solution/1900-1999/1953.Maximum Number of Weeks for Which You Can Work/README_EN.md

@@ -72,7 +72,15 @@ Thus, one milestone in project 0 will remain unfinished.
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Greedy
+
+We consider under what circumstances we cannot complete all stage tasks. If there is a project $i$ whose number of stage tasks is greater than the sum of the number of stage tasks of all other projects plus $1$, then we cannot complete all stage tasks. Otherwise, we can definitely complete all stage tasks by interlacing between different projects.
+
+We denote the sum of the number of stage tasks of all projects as $s$, and the maximum number of stage tasks as $mx$, then the sum of the number of stage tasks of all other projects is $rest = s - mx$.
+
+If $mx > rest + 1$, then we cannot complete all stage tasks, and at most we can complete $rest \times 2 + 1$ stage tasks. Otherwise, we can complete all stage tasks, the number is $s$.
+
+The time complexity is $O(n)$, where $n$ is the number of projects. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
@@ -126,6 +134,15 @@ func numberOfWeeks(milestones []int) int64 {
 }
 ```
 
+```ts
+function numberOfWeeks(milestones: number[]): number {
+    const mx = Math.max(...milestones);
+    const s = milestones.reduce((a, b) => a + b, 0);
+    const rest = s - mx;
+    return mx > rest + 1 ? rest * 2 + 1 : s;
+}
+```
+
 <!-- tabs:end -->
 
 <!-- end -->

solution/1900-1999/1953.Maximum Number of Weeks for Which You Can Work/Solution.ts

@@ -0,0 +1,6 @@
+function numberOfWeeks(milestones: number[]): number {
+    const mx = Math.max(...milestones);
+    const s = milestones.reduce((a, b) => a + b, 0);
+    const rest = s - mx;
+    return mx > rest + 1 ? rest * 2 + 1 : s;
+}