Problem

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

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

思路

• bit manipulation
  • 统计每个数字32位一共出现的次数
  • 然后将每一位长度大于 n/2的位记录下来，然后转换成数字

// 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
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;
}