[Leetcode] Majority Element

Majority Element

Given an array of size n, find the majority element. The majority element is the element, the appears more than times ? n/2 ? .

Assume that the array was non-empty and the majority element always exist in the array.

Problem Solving Ideas:

1. Use a map to record the number of occurrences of each number. If a count of a number is greater than N/2, the number is returned.

Class Solution {public:    int majorityelement (vector<int> &num) {        int len = Num.size ();        Map<int, int> count;        for (int i=0; i<len; i++) {            count[num[i]]++;            if (Count[num[i]] > LEN/2) {                return num[i];            }        }        return 0;}    };

2, for a sorted array, if the majority exists, the majority of the number must be the middle.

Class Solution {public:    int majorityelement (vector<int> &num) {        std::sort (Num.begin (), Num.end ());        return Num[num.size ()/2];    };

3, voting algorithm. Can think of as Daleitai, on stage that person if win and lose the same number of times, then step down, finally defeated his people on stage. The majority is certainly the last winner. This algorithm uses the least time.

Class Solution {public:    int majorityelement (vector<int> &num) {        int len = Num.size ();        if (len==0) {            return 0;        }        int candidate = num[0];        int count = 1;        for (int i=1; i<len; i++) {            if (num[i]==candidate) {                count++;            } else{                count--;            }            if (count==0) {                candidate = num[i];                count=1;            }        }        return candidate;}    ;

4, bit algorithm. Considering that each number can be represented by a 32-bit binary, the count of each digit binary is 1, if a bits count is greater than N/2, there must be a majority contribution. This approach is novel, although the speed is not comparable to voting algorithms, but pioneering thinking. To say here, the first method, defines Map<int, int> count, for each newly added key value, its value defaults to 0, but for the int array type, each array is initialized to a random value, so use the Memset function.

Class Solution {public:    int majorityelement (vector<int> &num) {        int len = Num.size ();        if (len==0) {            return 0;        }        int bitcount[32];        memset (bitcount, 0, 32*sizeof (int.));        for (int i=0, i<len; i++) {for            (int j=0; j<32; J + +) {                if (num[i]& (1<<j)) {  //J bit is 1                    bitcount[j]++        ;        }}} int result=0;        for (int i=0; i<32; i++) {            if (Bitcount[i] > LEN/2)         //I-bit is 1 count greater than half, there must be a contribution of the majority                result + = (int) Pow (2, i) ;        }        return result;}    ;

[Leetcode] Majority Element