leetcode 169.Majority Element

这道题虽然简单，但适合用来练习各种解法。《剑指offer》5.2节 面试题29与此题一样，并且给出了leetcode官方题解未给出的快速选择的解法。

方法一、用哈希表解决

class Solution {
public:
    int majorityElement(vector<int>& nums) {
        unordered_map<int,int> count;
        int majority = 0;
        int cnt = 0;
        for(auto num:nums){
            count[num]++;
            if(count[num] > cnt){
                cnt = count[num];
                majority = num;
            }
        }
        return majority;
    }
};

方法二、先排序再取numsn/2

class Solution {
public:
    int majorityElement(vector<int>& nums) {
        int len = nums.size();
        srand(time(0));
        quick_sort(nums,0,len-1);
        return nums[len/2];
    }

    int partition(vector<int>& nums,int left,int right){
        if(left >= right)
            return left;
        int random = rand()%(right - left +1);
        swap(nums[left],nums[left + random]);
        int pivot = nums[left];
        while(left < right){
            while(left<right && pivot < nums[right]) right--;
            nums[left] = nums[right];  
            while(left<right && nums[left]<=pivot) left++;
            nums[right] = nums[left];
        }
        nums[left] = pivot;
        return left;
    }
    void quick_sort(vector<int>& nums,int left,int right){
        if(left>=right) return;
        int pivot_pos = partition(nums,left,right);
        quick_sort(nums,left,pivot_pos-1);
        quick_sort(nums,pivot_pos+1,right);
    }
};

方法三、快速选择

由于方法二可知，问题等价于求数组排好序后的第n/2个元素。那么由快速排序算法的思想，其实不用把整个数组排序结束，就可以找到排好序后的第n/2个元素，也就是快速选择算法。

这种解法，leetcode官方没有给出。

class Solution {
public:
    int majorityElement(vector<int>& nums) {
        srand(time(0));
        return quick_select(nums,0,nums.size()-1,nums.size()/2);
    }

    int partition(vector<int>& nums,int left,int right){
        if(left >= right) return left;
        int random_pos = left + rand()%(right - left +1);
        swap(nums[left],nums[random_pos]);
        int pivot = nums[left];
        while(left<right){
            while(left<right && pivot <= nums[right]) right--;
            nums[left] = nums[right];
            while(left<right && nums[left] < pivot) left++;
            nums[right] = nums[left];
        }
        nums[left] = pivot;
        return left;
    }

    int quick_select(vector<int>& nums,int left,int right,int k){
        if(left>=right) return nums[left];
        int pivot_pos = partition(nums,left,right);
        if(pivot_pos == k)
            return nums[pivot_pos];
        else if(pivot_pos > k)
            return quick_select(nums,left,pivot_pos-1,k);
        else
            return quick_select(nums,pivot_pos+1,right,k);
    }
};

方法四、随机化方法

class Solution {
public:
    int majorityElement(vector<int>& nums) {
        srand(0);
        int len = nums.size();
        int random = 0; 
        int cnt = 0;
        while(true){
            random = rand()%len;
            cnt = 0;
            for(auto num:nums){
                if(num == nums[random]){
                    cnt++;
                    if(cnt >len/2)
                        return num;
                }
            }
        }
    }
};

方法五、分治法

class Solution {
public:
    int majorityElement(vector<int>& nums) {
        return divide_conquer(nums,0,nums.size()-1);
    }

    int divide_conquer(vector<int>& nums,int left,int right){
        if(left>=right)
            return nums[left];
        int mid = left + (right - left)/2;
        int majorityLeft = divide_conquer(nums,left,mid);
        int majorityRight = divide_conquer(nums,mid+1,right);
        if(majorityLeft == majorityRight)
            return majorityLeft;
        int cnt = 0;
        for(int i = left;i <= right;i++){
            if(nums[i] == majorityLeft){
                cnt++;
                if(cnt > (right - left +1)/2)
                    return majorityLeft;
            }
        }
        return majorityRight;
    }
};

方法六、Boyer-Moore投票法

《剑指offer》5.2节也给出了这种解法

class Solution {
public:
    int majorityElement(vector<int>& nums) {
        int candidate = nums[0];
        int count = 1;
        for(int i = 1;i < nums.size();i++){
            if(nums[i] == candidate){
                count++;
            }else{
                count--;
                if(count == 0)
                {
                    candidate = nums[i];
                    count = 1;
                }
            }
        }
        return candidate;
    }
};