🐱 Add Problem 169 of 5 Solutions

Author: sihgyu

src/0169.Majority-Element/README.md

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+# [169. Majority Element][title]
+
+## Description
+
+Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.
+
+You may assume that the array is non-empty and the majority element always exist in the array.
+**Example 1:**
+
+```
+Input: [3,2,3]
+Output: 3
+```
+
+**Example 2:**
+
+```
+Input: [2,2,1,1,1,2,2]
+Output: 2
+```
+
+**Tags:** Math, String
+
+## 题意
+>在一个数组除寻找出现次数超过一半的数
+
+## 题解
+
+### 思路1
+> 暴力2层循环
+
+```go
+func majorityElement(nums []int) int {
+major, count := nums[0], 0
+
+for _, v1 := range nums {
+tmpCount := 0
+for _, v2 := range nums {
+if v2 == v1 {
+tmpCount++
+}
+}
+if tmpCount > count {
+count = tmpCount
+major = v1
+}
+}
+
+return major
+}
+```
+
+### 思路2
+> Map 计数
+
+```go
+func majorityElement(nums []int) int {
+major, count := nums[0], 0
+majorMap := make(map[int]int)
+
+for _, v := range nums {
+majorMap[v]++
+}
+
+for i, v := range majorMap {
+if v > count {
+major = i
+count = v
+}
+}
+
+return major
+}
+```
+
+### 思路3
+> 先排序再循环判断每个数出现的次数
+
+```go
+func majorityElement4(nums []int) int {
+//将原数组排序
+nums = InsertSort(nums)
+
+major, count := nums[0], 0
+
+for _, v := range nums {
+if count == 0 {
+count++
+major = v
+} else if major == v {
+count++
+} else {
+count--
+}
+}
+
+return major
+}
+
+// 插入排序(Insert Sort)
+func InsertSort(arr []int) []int {
+var i, j, tmp int
+for i = 1; i < len(arr); i++ {
+tmp = arr[i]
+for j = i; j > 0 && arr[j-1] > tmp; j-- {
+arr[j] = arr[j-1]
+}
+arr[j] = tmp
+}
+return arr
+}
+```
+
+### 思路4
+> 利用分治分方法，借鉴了LeetCode某个大佬的分治解法，设计的非常精妙。
+
+
+```go
+func majorityElement(nums []int) int {
+if len(nums) == 1 {
+return nums[0]
+}
+n := len(nums)
+
+if n%2 == 1 {
+count := 1
+
+for i := 0; i < n-1; i++ {
+if nums[n-1] == nums[i] {
+count++
+}
+}
+if count > n/2 {
+return nums[n-1]
+}
+}
+numsTmp := []int{}
+
+for i := 0; i < n/2; i++ {
+if nums[2*i] == nums[2*i+1] {
+numsTmp = append(numsTmp, nums[2*i])
+}
+}
+
+return majorityElement(numsTmp)
+}
+```
+
+
+## 结语
+
+如果你同我一样热爱数据结构、算法、LeetCode，可以关注我 GitHub 上的 LeetCode 题解：[awesome-golang-leetcode][me]
+
+[title]: hthttps://leetcode.com/problems/majority-element/
+[me]: https://github.com/kylesliu/awesome-golang-leetcode

src/0169.Majority-Element/Solution.go

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+package Solution
+
+//	解法1： 循环计数，代码比较简单，LeetCode上的Case都是有Majority的，所以这里确实很多判断
+func majorityElement1(nums []int) int {
+major, count := nums[0], 0
+
+for _, v := range nums {
+if count == 0 {
+count++
+major = v
+} else if major == v {
+count++
+} else {
+count--
+}
+}
+
+return major
+}
+
+//	解法2: 2次循环
+func majorityElement2(nums []int) int {
+major, count := nums[0], 0
+
+for _, v1 := range nums {
+tmpCount := 0
+for _, v2 := range nums {
+if v2 == v1 {
+tmpCount++
+}
+}
+if tmpCount > count {
+count = tmpCount
+major = v1
+}
+}
+
+return major
+}
+
+//	解法3: Map
+func majorityElement3(nums []int) int {
+major, count := nums[0], 0
+majorMap := make(map[int]int)
+
+for _, v := range nums {
+majorMap[v]++
+}
+
+for i, v := range majorMap {
+if v > count {
+major = i
+count = v
+}
+}
+
+return major
+}
+
+//	解法4: 先排序再依次比对
+func majorityElement4(nums []int) int {
+//	将原数组排序
+nums = InsertSort(nums)
+
+major, count := nums[0], 0
+
+for _, v := range nums {
+if count == 0 {
+count++
+major = v
+} else if major == v {
+count++
+} else {
+count--
+}
+}
+
+return major
+}
+
+// 插入排序(Insert Sort)
+func InsertSort(arr []int) []int {
+var i, j, tmp int
+for i = 1; i < len(arr); i++ {
+tmp = arr[i]
+for j = i; j > 0 && arr[j-1] > tmp; j-- {
+arr[j] = arr[j-1]
+}
+arr[j] = tmp
+}
+return arr
+}
+
+//	解法5: 分治算法
+func majorityElement5(nums []int) int {
+if len(nums) == 1 {
+return nums[0]
+}
+n := len(nums)
+
+if n%2 == 1 {
+count := 1
+
+for i := 0; i < n-1; i++ {
+if nums[n-1] == nums[i] {
+count++
+}
+}
+if count > n/2 {
+return nums[n-1]
+}
+}
+numsTmp := []int{}
+
+for i := 0; i < n/2; i++ {
+if nums[2*i] == nums[2*i+1] {
+numsTmp = append(numsTmp, nums[2*i])
+}
+}
+
+return majorityElement1(numsTmp)
+}