Majority Element

Majority Element 

问题：

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.

思路：

　　超过半数的总会剩余

我的代码：

public class Solution {

    public int majorityElement(int[] num) {

        if(num == null || num.length == 0)  return -1;

        Stack<Integer> stack = new Stack<Integer>();

        for(int i = 0; i < num.length; i++)

        {

            if(stack.isEmpty()) stack.push(num[i]);

            else

            {

                int top = stack.peek();

                if(top == num[i])   stack.push(num[i]);

                else    stack.pop();

            }

        }

        return stack.pop();

    }

}

View Code

他人代码：

public class Solution {

    public int majorityElement(int[] num) {

        int maj=0;

        int count = 0;

        int n = num.length;

        for (int i = 0; i < n; i++){

            if (count == 0){

                maj = num[i];

                count++;

            }

            else if (num[i] == maj){

                count++;

                if (count > n/2) return maj;

            }

            else count--;

        }

        return maj;

    }

}

View Code

学习之处：

• 思路相似，我的方法空间复杂度O(n) 他人方法O(1) 
• maj 保存的是到目前扫描为止，发现的最大值，要不要更新最大值看count是否为0