feat: azl397985856#229

README.en.md

@@ -170,6 +170,7 @@ The data structures mainly includes:
 - [0209.minimum-size-subarray-sum](./problems/209.minimum-size-subarray-sum.md)
 - [0215.kth-largest-element-in-an-array](./problems/215.kth-largest-element-in-an-array.md) 🆕 
 - [0221.maximal-square](./problems/221.maximal-square.md) 🆕 
+- [0229.majority-element-ii](./problems/229.majority-element-ii.md) 🆕 
 - [0230.kth-smallest-element-in-a-bst](./problems/230.kth-smallest-element-in-a-bst.md)
 - [0236.lowest-common-ancestor-of-a-binary-tree](./problems/236.lowest-common-ancestor-of-a-binary-tree.md)🆕 
 - [0238.product-of-array-except-self](./problems/238.product-of-array-except-self.md)

README.md

@@ -189,6 +189,7 @@ leetcode 题解，记录自己的 leetcode 解题之路。
 - [0209.minimum-size-subarray-sum](./problems/209.minimum-size-subarray-sum.md) 
 - [0215.kth-largest-element-in-an-array](./problems/215.kth-largest-element-in-an-array.md) 🆕 
 - [0221.maximal-square](./problems/221.maximal-square.md) 🆕 
+- [0229.majority-element-ii](./problems/229.majority-element-ii.md) 🆕 
 - [0230.kth-smallest-element-in-a-bst](./problems/230.kth-smallest-element-in-a-bst.md)
 - [0236.lowest-common-ancestor-of-a-binary-tree](./problems/236.lowest-common-ancestor-of-a-binary-tree.md)🆕 
 - [0238.product-of-array-except-self](./problems/238.product-of-array-except-self.md)

problems/229.majority-element-ii.md

@@ -0,0 +1,167 @@
+
+## 题目地址
+https://leetcode.com/problems/majority-element-ii/description/
+
+## 题目描述
+
+```
+Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times.
+
+Note: The algorithm should run in linear time and in O(1) space.
+
+Example 1:
+
+Input: [3,2,3]
+Output: [3]
+Example 2:
+
+Input: [1,1,1,3,3,2,2,2]
+Output: [1,2]
+```
+
+## 思路
+
+这道题目和[169.majority-element](./169.majority-element.md) 很像。
+
+我们仍然可以采取同样的方法 - “摩尔投票法”， 具体的思路可以参考上面的题目。
+
+但是这里有一个不同的是这里的众数不再是超过`1 / 2`,而是超过`1 / 3`。
+题目也说明了，超过三分之一的有可能有多个（实际上就是0，1，2三种可能）。
+
+因此我们不能只用一个counter来解决了。 我们的思路是同时使用两个counter，其他思路和上一道题目一样。
+
+最后需要注意的是两个counter不一定都满足条件，这两个counter只是出现次数最多的两个数字。
+有可能不满足出现次数大于1/3， 因此最后我们需要进行过滤筛选。 
+
+这里画了一个图，大家可以感受一下：
+
+![229.majority-element-ii-1](../assets/problems/229.majority-element-ii-1.jpeg)
+
+![229.majority-element-ii-1](../assets/problems/229.majority-element-ii-2.jpeg)
+
+
+
+## 关键点解析
+
+- 摩尔投票法
+- 两个counter
+- 最后得到的只是出现次数最多的两个数字，有可能不满足出现次数大于1/3
+
+## 代码
+
+JavaScript代码：
+
+```js
+/*
+ * @lc app=leetcode id=229 lang=javascript
+ *
+ * [229] Majority Element II
+ */
+/**
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+var majorityElement = function(nums) {
+  const res = [];
+  const len = nums.length;
+  let n1 = null,
+    n2 = null,
+    cnt1 = 0,
+    cnt2 = 0;
+
+  for (let i = 0; i < len; i++) {
+    if (n1 === nums[i]) {
+      cnt1++;
+    } else if (n2 === nums[i]) {
+      cnt2++;
+    } else if (cnt1 === 0) {
+      n1 = nums[i];
+      cnt1++;
+    } else if (cnt2 === 0) {
+      n2 = nums[i];
+      cnt2++;
+    } else {
+      cnt1--;
+      cnt2--;
+    }
+  }
+
+  cnt1 = 0;
+  cnt2 = 0;
+
+  for (let i = 0; i < len; i++) {
+    if (n1 === nums[i]) {
+      cnt1++;
+    } else if (n2 === nums[i]) {
+      cnt2++;
+    }
+  }
+
+  if (cnt1 > (len / 3) >>> 0) {
+    res.push(n1);
+  }
+  if (cnt2 > (len / 3) >>> 0) {
+    res.push(n2);
+  }
+
+  return res;
+};
+
+```
+
+Java代码：
+
+```java
+/*
+ * @lc app=leetcode id=229 lang=java
+ *
+ * [229] Majority Element II
+ */
+class Solution {
+    public List<Integer> majorityElement(int[] nums) {
+        List<Integer> res = new ArrayList<Integer>();
+        if (nums == null || nums.length == 0)
+            return res;
+        int n1 = nums[0], n2 = nums[0], cnt1 = 0, cnt2 = 0, len = nums.length;
+        for (int i = 0; i < len; i++) {
+            if (nums[i] == n1)
+                cnt1++;
+            else if (nums[i] == n2)
+                cnt2++;
+            else if (cnt1 == 0) {
+                n1 = nums[i];
+                cnt1 = 1;
+            } else if (cnt2 == 0) {
+                n2 = nums[i];
+                cnt2 = 1;
+            } else {
+                cnt1--;
+                cnt2--;
+            }
+        }
+        cnt1 = 0;
+        cnt2 = 0;
+        for (int i = 0; i < len; i++) {
+            if (nums[i] == n1)
+                cnt1++;
+            else if (nums[i] == n2)
+                cnt2++;
+        }
+        if (cnt1 > len / 3)
+            res.add(n1);
+        if (cnt2 > len / 3 && n1 != n2)
+            res.add(n2);
+        return res;
+    }
+}
+
+```
+
+
+## 扩展
+
+如果题目中3变成了k，怎么解决？
+
+大家可以自己思考一下，我这里给一个参考链接：https://leetcode.com/problems/majority-element-ii/discuss/63500/JAVA-Easy-Version-To-Understand!!!!!!!!!!!!/64925
+
+这个实现说实话不是很好，大家可以优化一下。