From 5324e07e527a23d32835e020e217de270ddd2094 Mon Sep 17 00:00:00 2001
From: "pengpeng.chen" <pengpeng.chen@ximalaya.com>
Date: Sat, 10 Feb 2024 09:14:37 +0800
Subject: [PATCH] feat: 0209

---
 docs/interview/experiences/22leetcodeB.md     | 30 ++++++++++++++++
 .../experiences/practise/202401/0126.js       | 36 +++++++++++++++++++
 2 files changed, 66 insertions(+)
 create mode 100644 docs/interview/experiences/practise/202401/0126.js

diff --git a/docs/interview/experiences/22leetcodeB.md b/docs/interview/experiences/22leetcodeB.md
index ce05459..e719d08 100644
--- a/docs/interview/experiences/22leetcodeB.md
+++ b/docs/interview/experiences/22leetcodeB.md
@@ -72,6 +72,7 @@ nav:
  * 48.滑动窗口最大值
  * 49.除自身以外数组的乘积
  * 50.旋转图像
+ * 51.分发饼干
  */
 ```
 
@@ -959,3 +960,32 @@ mul([1, 2, 3, 4]);
 ```
 
 ## 50.旋转图像
+
+## [51.分发饼干](https://leetcode.cn/problems/assign-cookies/description/)
+
+> 尽量用大饼干喂胃口大的孩子
+
+```js
+/**
+输入: g = [1,2,3], s = [1,1]
+输出: 1
+解释:
+你有三个孩子和两块小饼干，3个孩子的胃口值分别是：1,2,3。
+你的目标是尽可能满足越多数量的孩子，并输出这个最大数值。
+*/
+var findContentChildren = function (g, s) {
+  s.sort((a, b) => a - b);
+  g.sort((a, b) => a - b);
+  let count = 0;
+  let sLen = s.length - 1;
+  let gLen = g.length - 1;
+  for (let i = gLen; i >= 0; i--) {
+    let item = g[i];
+    if (sLen >= 0 && s[sLen] >= item) {
+      count++;
+      sLen--;
+    }
+  }
+  return count;
+};
+```
diff --git a/docs/interview/experiences/practise/202401/0126.js b/docs/interview/experiences/practise/202401/0126.js
new file mode 100644
index 0000000..07d408f
--- /dev/null
+++ b/docs/interview/experiences/practise/202401/0126.js
@@ -0,0 +1,36 @@
+/**
+ * 1.分发饼干
+ * 2.摆动序列
+ */
+
+var wiggleMaxLength = function (nums) {
+  let max = 0;
+  let dp = new Array(nums.length + 1).fill(0);
+  dp[0] = 1;
+  for (let i = 1; i <= nums.length; i++) {}
+};
+
+就只有8号上午的时间;
+// 大姨二姨舅舅 上午走完 4个
+// 二姑小姑 中午 2个 总共6个 小朋友俩个红包
+/**
+ * 
+输入: g = [1,2,3], s = [1,1]
+输出: 1
+解释: 
+你有三个孩子和两块小饼干，3个孩子的胃口值分别是：1,2,3。
+你的目标是尽可能满足越多数量的孩子，并输出这个最大数值。
+*/
+var findContentChildren = function (g, s) {
+  s.sort((a, b) => a - b);
+  g.sort((a, b) => a - b);
+  let count = 0;
+  let sLen = s.length - 1;
+  for (let item of g) {
+    if (sLen >= 0 && s[sLen] >= item) {
+      count++;
+      sLen--;
+    }
+  }
+  return count;
+};