169 Majority Element [LeetCode Java实现]

题目链接：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.
 *
 */
public class MajorityElement {

//	40 / 40 test cases passed.
//	Status: Accepted
//	Runtime: 253 ms
//	Submitted: 1 minute ago

	//时间复杂度为 O(n), 空间复杂度为 O(1)

	//因为最多的那个元素占一半及以上， 故能够和别的元素相互抵消，最后剩下的未抵消掉的元素为所求
	//将数组分成三块：num[0...i]为 归并的 还未抵消的数组。 num[i+1...j－1]空暇区， num[j...num.length-1]为等待抵消的数组
	//抵消规则：若num[i] = num[j] 则把num[j]归入num[0...i+1]数组中
	//		 若num[i] != num[j] 则把num[i] 抵消掉 然后 i－－

    static int majorityElement(int[] num) {

    	int i = -1;

    	for (int j = 0; j < num.length; j++) {

    		if(i == -1) num[++i] = num[j];
        	else {
            	if(num[j] == num[i]) num[ ++i] = num[j];
    			else i --;
			}

		}
    	return num[0];
    }
	public static void main(String[] args) {

		System.out.println(majorityElement(new int[]{4}));
		System.out.println(majorityElement(new int[]{4, 4, 5}));
		System.out.println(majorityElement(new int[]{4, 3, 3}));
		System.out.println(majorityElement(new int[]{4, 3, 4}));
	}

}