标签:style blog java color cti for
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?
Analysis: 注意那个single one的个数可能比3少,也可能比3多。所以想法就是先sorting,然后从小到大遍历每个元素,前后元素相同时counter++,不相同时看counter是否为3,不是的话就返回前一个元素,是的话counter重新置1. 特殊情况在于single one在最后,所以如果循环执行完了函数都还没有return,说明single one是最后的元素。要注意,对于这种需要返回值的function, 不能所有的return都在if语句里
1 public class Solution { 2 public int singleNumber(int[] A) { 3 java.util.Arrays.sort(A); 4 int count = 1; 5 int i; 6 for (i = 1; i < A.length; i++) { 7 if (A[i] == A[i-1]) { 8 count++; 9 } 10 else { 11 if (count != 3) return A[i-1]; 12 count = 1; 13 } 14 } 15 return A[i-1]; 16 } 17 }
Leetcode: Single Number II,布布扣,bubuko.com
标签:style blog java color cti for
原文地址:http://www.cnblogs.com/EdwardLiu/p/3798586.html