代码面试题：Majority Number II

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#面试题

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本文介绍了一种通过三三抵销的方法来找到数组中出现超过数组长度三分之一的元素，包括思路解析和具体实现代码。

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Given an array of integers, the majority number is the number that occurs more than 1/3 of the size of the array.

Find it.

Note

There is only one majority number in the array

Example

For [1, 2, 1, 2, 1, 3, 3] return 1

思路：三三抵销，同时记录剩下的值各自出现的次数，出现次数多的值为答案。

public int majorityNumber2(int[] A){
    	if (A == null || A.length == 0) {
    		return -1;
    	}
    	
    	int candidate1 = 0;
    	int candidate2 = 0;
    	int count1 = 0;
    	int count2 = 0;
    	for (int i = 0; i < A.length; i++) {
    		if (count1 == 0) {
    			candidate1 = A[i];
    		}
    		if (count2 == 0 && A[i] != candidate1) {
    			candidate2 = A[i];
    		}
    		if (A[i] != candidate1 && A[i] != candidate2 
    				&& count1 != 0 && count2 != 0) {
    			count1--;
    			count2--;
    		} 
    		if (A[i] == candidate1) {
    			count1++;
    		}
    		if (A[i] == candidate2) {
    			count2++;
    		}
    	}
    	
    	if (count1 > count2) {
    		return candidate1;
    	} else {
    		return candidate2;
    	}
    }