169. Majority Element

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.

这道题的naive解法是O(n2)，两个for循环，挨个统计数字出现的个数，超过n/2就返回，提交完之后第42个测试实例超时了，只能想复杂度更低的算法。

这里要隆重介绍一下O(n)复杂度的 Boyer–Moore majority vote algorithm。这是一个专门统计序列中出现频率最高的字符的算法，伪代码如下：

Initialize an element m and a counter i with i = 0For each element x of the input sequence:If i = 0, then assign m = x and i = 1else if m = x, then assign i = i + 1else assign i = i − 1Return m

转换成这道题 java 的解法就是：

public class Solution {    public int majorityElement(int[] nums) {        int i = 0, count = 0, majority = 0;        while (i < nums.length) {            if (count == 0) {                majority = nums[i];                count ++;            } else if (majority == nums[i]) {                count ++;            } else {                count --;            }            i++;        }        return majority;    }}

实在是巧妙。其实这道题还有很多其他的解法，比如sorting, hashmap, moore voting, bit manipulation。具体的代码实现如下：

// Sortingpublic 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 algorithmpublic 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;}

最后一个 Bit manipulation 非常巧妙，因为序列中的数都是 int 型（32位）。分别统计序列中每一个数每一位出现的次数，大于n/2的位即属于出现次数大于n/2的那个数，把属于的位合并，得到最终结果，思想是Divide and Conquer。