Single Number II

标签：leetcode

Given an array of integers, every element appears three times except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

题目解析：数组都是3个3个出现，找出唯一的数字，这个是在2个2个出现的升级版，2个2个直接异或即可

解题思路：先用快排将数组排序，然后维护一个times和result字段，另其初始为2，如果下一个数跟前一个数相等，则times--，当times为01说明3个已经过去，将times重新赋值为3，result=nums[i]，按照这种方式遍历一边即可，这里的这个小技巧来自剑指offer面试题29，下面上AC代码

 public int singleNumber(int[] array) {
        // 安全性检查 数组是否符合3*n+1;
		float len = array.length;
		if(len==1.0)
		return array[0];
		if ((len - 1) % 3 != 0 || len < 4)
			return -1;

		quick_sort(array, 0, array.length - 1);
		int FindNum = array[0];
		int times = 2;
		for (int i = 1; i < len; i++) {
			if (times == 0) {
				// 换下一个数，times置为3重新开始
				FindNum = array[i];
				times = 3;
			}
			if (FindNum == array[i])
				times--;
		}

		return FindNum;
    }
    // 快速排序
	private static void quick_sort(int[] arr, int low, int high) {
		// 解决和合并
		if (low <= high) {
			int mid = partition(arr, low, high);
			// 递归
			quick_sort(arr, low, mid - 1);
			quick_sort(arr, mid + 1, high);
		}

	}

	private static int partition(int[] arr, int low, int high) {
		// 分解
		int pivot = arr[high];
		int i = low - 1;
		int temp;
		for (int j = low; j < high; j++) {

			if (arr[j] < pivot) {
				i++;
				temp = arr[i];
				arr[i] = arr[j];
				arr[j] = temp;
			}
		}
		// 交换中间元素和privot
		temp = arr[i + 1];
		arr[i + 1] = arr[high];
		arr[high] = temp;
		return i + 1;

	}

Single Number II

标签：leetcode

原文地址：http://blog.csdn.net/xsf50717/article/details/47135973