169. Majority Element

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

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

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

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@jinlibao的分治解法,设计的非常精妙。

流程:

    1. 如果数组长度是奇数,找出与末尾数相等的总个数,个数大于n/2直接return;
    1. 去掉末尾数,两两比较相等则放入新数组;
    1. 递归调用,传入新数组;

这个算法妙就妙在抓住了4个数中必然有3个数相同,两两比较相等只会留下1个数就是结果,所以只要用分治思路将数组分到4个长度就能得到结果。

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

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