标签：edits edit turn case line can anim blank ges
169. 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.
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
public class Solution { public int majorityElement(int[] nums) { int candidate = 0; int count = 0; for (int num : nums) { if (count == 0) { candidate = num; count = 1; } else if (candidate == num) { count++; } else { count--; } } return candidate; } }
229. Majority Element II
Given an integer array of size n, find all elements that appear more than ? n/3 ? times. The algorithm should run in linear time and in O(1) space.
Hint:
Do you have a better hint? Suggest it!
public class Solution { public List<Integer> majorityElement(int[] nums) { int candidate1 = 0, candidate2 = 0; int count1 = 0, count2 = 0; for (int num : nums) { if (candidate1 == num) { count1++; } else if (candidate2 == num) { count2++; } else if (count1 == 0) { candidate1 = num; count1 = 1; } else if (count2 == 0) { candidate2 = num; count2 = 1; } else { count1--; count2--; } } List<Integer> result = new ArrayList<Integer>(); int length = nums.length; /* if (count1 == 0 && count2 == 0) { return result; } else if (count1 > 0 && count2 == 0) { result.add(candidate1); return result; } else if (count2 > 0 && count1 == 0) { result.add(candidate2); return result; }*/ count1 = 0; count2 = 0; for (int num : nums) { if (num == candidate1) { count1++; } else if (num == candidate2) { count2++; } } if (count1 > length / 3) { result.add(candidate1); } if (count2 > length / 3) { result.add(candidate2); } return result; } }
标签：edits edit turn case line can anim blank ges
原文地址：http://www.cnblogs.com/Gryffin/p/6238594.html
