LeetCode-169.229. Majority Element II (JAVA)主要元素
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169. Majority Element
Given an array of size n, find the majority element. The majority element is the element that appearsmore than⌊ n/2 ⌋
times.
You may assume that the array is non-empty and the majority element always exist in the array.
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
从一个数组中找出出现半数以上的元素。(数组肯定存在这样的元素)
排序:
返回中间元素
// Sortingpublic int majorityElement(int[] nums) { Arrays.sort(nums); return nums[nums.length/2];}Hashtable(jdk8语法)
public int majorityElement(int[] nums) { Map<Integer, Integer> map = new HashMap<Integer, Integer>(); int ret=0; for (int num: nums) { map.put(num, map.getOrDefault(num, 0)+1); if (map.get(num)>nums.length/2) { ret = num; break; } } return ret;}}Moore voting algorithm
每次都找出一对不同的元素,从数组中删掉,直到数组为空或只有一种元素。 不难证明,如果存在元素e出现频率超过半数,那么数组中最后剩下的就只有e。
当然,最后剩下的元素也可能并没有出现半数以上。比如说数组是[1, 2, 3],最后剩下的3显然只出现了1次,并不到半数。排除这种false positive情况的方法也很简单,只要保存下原始数组,最后扫描一遍验证一下就可以了。
public int majorityElement(int[] nums) { int count = 0, ret = 0; for (int num : nums) { if (count == 0) ret = num; if (num != ret) count--; else count++; } return ret; }
229. Majority Element II
Given an integer array of size n, find all elements that appear more than⌊ n/3 ⌋
times. The algorithm should run in linear time and in O(1) space.
借鉴上一题的 Moore voting algorithm
给定一个整型数组,找到所有主元素,它在数组中的出现次数严格大于数组元素个数的三分之一。
三三抵消,最后会剩下两个candidates,但是注意此时不是谁占多数谁是最终结果,反例[1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 4, 4] 三三抵消后剩下 [1, 4,4] 4数量占优,但结果应该是1,所以三三抵消后,再loop一遍找1和4谁数量超过了len(nums)/3
public List<Integer> majorityElement(int[] nums) {List<Integer> ret = new ArrayList<>();int n = nums.length;int candidate1 = 0, candidate2 = 0, counter1 = 0, counter2 = 0;for (int i : nums) {if (candidate1 == i) {counter1++;} else if (candidate2 == i) {counter2++;} else if (counter1 == 0) {candidate1 = i;counter1 = 1;} else if (counter2 == 0) {candidate2 = i;counter2 = 1;} else {counter1--;counter2--;}}// 统计数量多的,最后剩下元素counter1 = 0;counter2 = 0;for (int num : nums) {if (num == candidate1)counter1++;else if (num == candidate2)counter2++;}if (counter1 > n / 3)ret.add(candidate1);// 不能直接else,可能为空集if (counter2 > n / 3)ret.add(candidate2);return ret;}
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