LeetCode & Q169-Majority Element-Easy

Array Divide and Conquer Bit Manipulation

Description:

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.

我的第一种解法特别蠢，想着个数超过n/2，肯定有连续的，找出连续的就行了，当然其他元素也可以连续....真的怀疑自己脑子少根筋...

好一点的解法要引入count，至于这个变量记录什么是很讲究的，为了使时间复杂度尽量降，就用一个count记录整个数组的遍历过程。count初始化为0，当当前遍历的数字与初定的major相同，count++，否则count--。major的值随count变为0后转成下一个数。其实这就是最大投票算法。

my Solution:

public class Solution {
    public int majorityElement(int[] nums) {
        int major = nums[0];
        int count = 0;
        for (int num : nums) {
            if (count == 0) {
                major = num;
                count++;
            } else if (major == num) {
                count++;
            } else {
                count--;
            }
        }
        return major;
    }
}

在Discuss里看到有大牛一题多解了，此处膜拜一下

// Sorting
public int majorityElement1(int[] nums) {
    Arrays.sort(nums);
    return nums[nums.length/2];
}

// Hashtable
public int majorityElement2(int[] nums) {
    Map<Integer, Integer> myMap = new HashMap<Integer, Integer>();
    //Hashtable<Integer, Integer> myMap = new Hashtable<Integer, Integer>();
    int ret=0;
    for (int num: nums) {
        if (!myMap.containsKey(num))
            myMap.put(num, 1);
        else
            myMap.put(num, myMap.get(num)+1);
        if (myMap.get(num)>nums.length/2) {
            ret = num;
            break;
        }
    }
    return ret;
}

// Moore voting algorithm  就是题主的解法
public int majorityElement3(int[] nums) {
    int count=0, ret = 0;
    for (int num: nums) {
        if (count==0)
            ret = num;
        if (num!=ret)
            count--;
        else
            count++;
    }
    return ret;
}

// Bit manipulation
public int majorityElement(int[] nums) {
    int[] bit = new int[32];
    for (int num: nums)
        for (int i=0; i<32; i++)
            if ((num>>(31-i) & 1) == 1)
                bit[i]++;
    int ret=0;
    for (int i=0; i<32; i++) {
        bit[i]=bit[i]>nums.length/2?1:0;
        ret += bit[i]*(1<<(31-i));
    }
    return ret;
}