《leetCode》：Majority Element II

题目

Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times.
The algorithm should run in linear time and in O(1) space.

思路

The idea is to split the array into three parts according to the selected pivot: left, middle and right.

Let's say we have two indices m and n, so that all elements in [0 ... m-1] are less than the pivot,
elements in [m...n] are equal to the pivot (two ends inclusive), [n+1 ... end] contains elements greater than the pivot.
Then there are some facts:

if n - m + 1 >= 1 + nums.size()/3, nums[m] must be added to the results.

If m - 1 < 1 + nums.size()/3, we can simply abandon the left part, otherwise we have to consider it.

if nums.size()-n < 1+nums.size()/3, the right part can be dropped, otherwise it has to be checked.

Ideally, we can drop about 1/3 of the array each time,
so the running time is something like: 1 + 2/3 + (2/3)*(2/3) + ... = (1-(2/3)^k)/(1-2/3), O(n).

A big advantage of this algorithm is that we can simply apply it to 1/4,1/5 ...

/**
* Return an array of size *returnSize.
* Note: The returned array must be malloced, assume caller calls free().
*/
int *res=NULL;
int indexRes=0;

void swap(int *a,int *b){
int temp=*a;
*a=*b;
*b=temp;
}
void majorityElementHelper(int *nums,int left,int right,int len){
if(nums==NULL||right<left){
return;
}
//选取mid位置的值作为pivot值
int mid=(left+right)/2;
int val=nums[mid];

int m=left;//用来指示 调整后的结果中，m左边的值都是小于val的。
int n=left;//用来指示  调整后的结果中，m~n中的值都是等于val的。n~right都是大于val的
swap(nums+left,nums+mid);//将pivot放入nums第一个位置
int i=left+1;
while(i<=right){
if(nums[i]<val){
swap(&nums[m++],&nums[i]);//将小于val的值放入m下标的左边
swap(&nums[++n],&nums[i++]);//使得n总是指向最右边的等于val的下标
}
else if(nums[i]==val){
swap(&nums[++n],&nums[i++]);
}
else{
i++;
}
}
if(n-m+1>=len){//符合条件
res[indexRes++]=nums
;
}
if(m-left>=len){//小于val的长度大于len，因此可能有解
majorityElementHelper(nums,left,m-1,len);
}
if(right-n>=len){//大于val的长度大于len，因此可能有解
majorityElementHelper(nums,n+1,right,len);
}
}

int* majorityElement(int* nums, int numsSize, int* returnSize) {
if(nums==NULL||numsSize<1){
*returnSize=0;
return NULL;
}
int maxLen=3;//最多只有三个解
res=(int *)malloc(maxLen*sizeof(int));
if(res==NULL){
exit(EXIT_FAILURE);
}
indexRes=0;
majorityElementHelper(nums,0,numsSize-1,(numsSize/3)+1);
*returnSize=indexRes;
return res;
}