LeetCode169. Majority Element

题目

Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.

You may assume that the array is non-empty and the majority element always exist in the array.

答案1

第一反应用HashMap，时间复杂度O(n)，空间复杂度O(n)，不推荐。

public int majorityElement(int[] nums) {
    HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
    for (int i : nums) {
        if (map.containsKey(i)) map.put(i, map.get(i) + 1);
        else map.put(i, 1);
    }
    Iterator iterator = map.entrySet().iterator();
    while (iterator.hasNext()) {
        Map.Entry entry = (Map.Entry) iterator.next();
        if ((int) entry.getValue() > nums.length / 2) return (int) entry.getKey();
    }
    return 0;
}

答案2

排序后边遍历边计数，又出现次数超过 n / 2 的就return，时间复杂度O(nlogn)，空间复杂度O(1)。

public int majorityElement(int[] nums) {
    Arrays.sort(nums);
    int times = 1;
    for (int i = 0; i < nums.length - 1; i++){
        if (nums[i + 1] == nums[i]) {
            times++;
            if (times > nums.length / 2) return nums[i];
        } else times = 1;
    }
    return nums[0];
}

答案3

如果出现次数超过 n / 2 ，则第 n / 2 项一定是目标数字。

public int majorityElement(int[] nums) {
    Arrays.sort(nums);
    return nums[nums.length/2];
}