共计 1123 个字符，预计需要花费 3 分钟才能阅读完成。
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]
难度：medium
题目：给定一整数数组，找出其元素出现次数大于其三分之一长度的所有元素。注意：算法时间复杂度 O(n), 空间复杂度为 O(1)
思路：投票法。1 找出三个不同的候选人，各消除一票，如果票数为 0 退出候选人队列。2 最后对剩下的候选人重新统计票数
Runtime: 2 ms, faster than 97.44% of Java online submissions for Majority Element II.Memory Usage: 39.8 MB, less than 29.68% of Java online submissions for Majority Element II.
class Solution {
public List<Integer> majorityElement(int[] nums) {
List<Integer> result = new ArrayList<>();
if (null == nums) {
return result;
}
int[] elems = new int[2];
int[] count = new int[2];
for (int i = 0; i < nums.length; i++) {
if (elems[0] == nums[i]) {
count[0]++;
} else if (elems[1] == nums[i]) {
count[1]++;
} else {
if (count[0] > 0 && count[1] > 0) {
count[0]–;
count[1]–;
} else if (0 == count[0]) {
elems[0] = nums[i];
count[0] = 1;
} else if (0 == count[1]) {
elems[1] = nums[i];
count[1] = 1;
}
}
}
int c0 = 0, c1 = 0;
for (int i = 0; i < nums.length; i++) {
if (count[0] > 0 && nums[i] == elems[0]) {
c0++;
}
if (count[1] > 0 && nums[i] == elems[1]) {
c1++;
}
}
if (c0 > nums.length / 3) {
result.add(elems[0]);
}
if (c1 > nums.length / 3) {
result.add(elems[1]);
}
return result;
}
}
