Majority Element II——LeetCode

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.

题目大意：给定一个大小为n的数组，找出Majority元素，满足出现次数大于

⌊ n/3 ⌋

次。

解题思路：大于

⌊ n/3 ⌋

次，那么应该最多有两个Majority元素，之前有道题是找出 大于

⌊ n/2 ⌋

次Majority元素，这道题有点类似，稍微复杂一点，还是利用投票算法，count作为对元素的计数，候选元素等于当前元素则count+1，否则减1，count等于0则更换候选元素。

public List<Integer> majorityElement(int[] nums) {
List<Integer> res = new ArrayList<>();
if (nums == null || nums.length == 0) {
return res;
}
int can1 = nums[0], can2 = nums[0], cnt1 = 1, cnt2 = 0;
for (int i = 1; i < nums.length; i++) {
int num = nums[i];
if ((cnt1 == 0 && num != can2)) {
can1 = num;
} else if (cnt2 == 0 && num != can1) {
can2 = num;
}
if (num == can1) {
cnt1++;
} else if (num == can2) {
cnt2++;
} else {
cnt1--;
cnt2--;
}
cnt1 = cnt1 < 0 ? 0 : cnt1;
cnt2 = cnt2 < 0 ? 0 : cnt2;
}
cnt1 = 0;
cnt2 = 0;
for (int num : nums) {
if (num == can1) {
cnt1++;
} else if (num == can2) {
cnt2++;
}
}
if (cnt1 > nums.length / 3)
res.add(can1);
if (cnt2 > nums.length / 3)
res.add(can2);
return res;
}