0229. Majority Element II (M)

Majority Element II (M)

題目

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]

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題意

找到數組中全部出現次數大於⌊ n/3 ⌋的元素。

思路

限制時間爲O(N)、空間爲O(1)，所以不能用Hash或者排序。使用 169. Majority Element (E) 中提到的摩爾投票法 Boyer-Moore Majority Vote。求全部出現次數大於⌊ n/3 ⌋的元素，用反證法很容易證實這種元素最多隻有兩個，因此能夠先用摩爾投票法得到兩個候選元素，再從新遍歷數組統計候選元素出現的次數，將知足條件的加入到結果集中。

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代碼實現

Java

class Solution {
    public List<Integer> majorityElement(int[] nums) {
        List<Integer> ans = new ArrayList<>();
        int a = 0, b = 0;
        int countA = 0, countB = 0;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == a) {
                countA++;
            } else if (nums[i] == b) {
                countB++;
            } else if (countA == 0) {
                a = nums[i];
                countA = 1;
            } else if (countB == 0) {
                b = nums[i];
                countB = 1;
            } else {
                countA--;
                countB--;
            }
        }
        countA = 0;
        countB = 0;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == a) {
                countA++;
            } else if (nums[i] == b) {
                countB++;
            }
        }
        if (countA > nums.length / 3) {
            ans.add(a);
        }
        if (countB > nums.length / 3) {
            ans.add(b);
        }
        return ans;
    }
}

JavaScript

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var majorityElement = function (nums) {
  let ans = []
  let a = 0, b = 0, countA = 0, countB = 0

  for (let num of nums) {
    if (num === a) {
      countA++
    } else if (num === b) {
      countB++
    } else if (countA === 0) {
      a = num
      countA = 1
    } else if (countB === 0) {
      b = num
      countB = 1
    } else {
      countA--
      countB--
    }
  }
  
  countA = 0, countB = 0
  for (let num of nums) {
    if (num === a) {
      countA++
    } else if (num === b) {
      countB++
    }
  }

  if (countA > Math.trunc(nums.length / 3)) ans.push(a)
  if (countB > Math.trunc(nums.length / 3)) ans.push(b)
  
  return ans
}