Majority Element II

Total Accepted: 3172 Total Submissions: 14746

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.

[思路]

[REFERENCE] http://bookshadow.com/weblog/2015/06/29/leetcode-majority-element-ii/

观察可知，数组中至多可能会有2个出现次数超过 ⌊ n/3 ⌋ 的众数

记变量n1, n2为候选众数； c1, c2为它们对应的出现次数

遍历数组，记当前数字为num

若num与n1或n2相同，则将其对应的出现次数加1

否则，若c1或c2为0，则将其置为1，对应的候选众数置为num

否则，将c1与c2分别减1

最后，再统计一次候选众数在数组中出现的次数，若满足要求，则返回之。

[CODE]

public class Solution {public List<Integer> majorityElement(int[] nums) {// 1, 2List<Integer> res = new ArrayList<>();if(nums==null || nums.length==0) return res;if(nums.length==1) {res.add(nums[0]);return res;}int m1 = nums[0];int m2 = 0;int c1 = 1;int c2 = 0;for(int i=1; i<nums.length; i++) {int x = nums[i];if(x==m1) ++c1;else if(x==m2) ++c2;else if(c1==0) {m1 = x;c1 = 1;} else if(c2==0) {m2 = x;c2 = 1;} else {--c1; --c2;}}c1 = 0; c2 = 0;for(int i=0; i<nums.length; i++) {if(m1 == nums[i]) ++c1;else if(m2 == nums[i]) ++c2;}if(c1>nums.length/3) res.add(m1);if(c2>nums.length/3) res.add(m2);return res;}
}