05 August 2008
Given an integer array of size n, find all elements that appear more than 1
⌊ n/3 ⌋
Note: The algorithm should run in linear time and in O(1) space.
Example 1:
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Input: [3,2,3]
Output: [3]
Example 2:
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Input: [1,1,1,3,3,2,2,2]
Output: [1,2]
跟169 Major Element相比，这道题是要求返回所出现次数超过1/3的元素，另外还规定了线性时间复杂度和常数时间，所以想利用数据结构来统计不可行；因为是出现次数大于1/3，所以众数最多是两个，最少是没有。做法是两次遍历，第一次遍历找出两个众数作为备选，第二次遍历利用投票方法确认这两个众数是否合格。
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class Solution {
public List<Integer> majorityElement(int[] nums) {
List<Integer> result = new ArrayList<>();
if (nums == null || nums.length == 0) return result;
int candidate1 = 0, candidate2 = 0, count1 = 0, count2 = 0;
for (int num : nums) {//利用投票法则，至少有两个数的出现次数是多于其它数的
if (num == candidate1) {
count1++;
} else if (num == candidate2) {
count2++;
} else if (count1 == 0) {
candidate1 = num;
count1 = 1;
} else if (count2 == 0) {
candidate2 = num;
count2 = 1;
} else {
count1--;
count2--;
}
}
//重新计算一下两个备选众数的实际出现次数，看是否超过1/3
count1 = 0;
count2 = 0;
for (int num : nums) {
if (candidate1 == num) {
count1++;
} else if (candidate2 == num) { //else if 表示candidate1和candidate2不会是一个数
count2++;
}
}
if (count1 > nums.length/3) result.add(candidate1);
if (count2 > nums.length/3) result.add(candidate2);
return result;
}
}