169. Majority Element Java Solutin

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.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

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public class Solution {
    public int majorityElement(int[] nums) {
        if(nums == null) return 0;
        Map<Integer,Integer> hm = new HashMap<Integer ,Integer>();
        for(int i= 0;i< nums.length;i++){
            if(hm.containsKey(nums[i])) hm.put(nums[i],hm.get(nums[i]) + 1);
            else{
                hm.put(nums[i],1);
            }
        }
        int halflen = nums.length/2;
        int res = 0;
        for(int key : hm.keySet()){
            if(hm.get(key) > halflen){
                res = key;
                break;
            }
        }
        return res;
    }
}

方法2： 排序后，取中间的元素就是所求值。

方法3：转自：http://blog.csdn.net/feliciafay/article/details/45006661

从头到尾扫描一遍数组，记录当前的majority element的count。

// 259ms
public class Solution {
    public int majorityElement(int[] num) {
        if(num.length < 3) return num[0];
        int majority = num[0];
        int count = 1;
        //1,1,1,1,2,1,3,1,2,2,2,2,2,2,
        for (int i = 1; i < num.length; ++i) {
            if (count == 0) {
                majority = num[i];
                ++count;
            } else if (num[i] == majority)  {
                ++count;
            } else {
                --count;
            }
        }
        return majority;
    }
}