LeetCode229 majorityelementii Java

<p> title: </p><p><span style= "Color:rgb (51, 51, 51); font-family: ' Helvetica Neue ', Helvetica, Arial, Sans-serif; font-size:14px; line-height:30px; " >given an integer array of Size </span><span style= "Box-sizing:border-box; Color:rgb (51, 51, 51); font-family: ' Helvetica Neue ', Helvetica, Arial, Sans-serif; font-size:14px; line-height:30px; " >n</span><span style= "Color:rgb (51, 51, 51); font-family: ' Helvetica Neue ', Helvetica, Arial, Sans-serif; font-size:14px; line-height:30px; " Elements, find all, this appear more Than </span><code style= "Box-sizing:border-box; Font-family:menlo, Monaco, Consolas, ' Courier New ', monospace; font-size:12.6000003814697px; PADDING:2PX 4px; Color:rgb (199, 37, 78); border-radius:4px; line-height:30px; Background-color:rgb (249, 242, 244); " ? N/3 </code><span style= "Color:rgb (51, 51, 51); font-family: ' Helvetica Neue ', Helvetica, Arial, Sans-serif; font-size:14px; Line-height:30px; " > times. The algorithm should run in linear time and in O (1) space.</span></p><p><span style= "Color:rgb (51, 5 1, 51); font-family: ' Helvetica Neue ', Helvetica, Arial, Sans-serif; font-size:14px; line-height:30px; " > The number of occurrences is greater than one-third of the length of the array, so there are only 2 such elements, the topic requires linear time complexity and constant space complexity, so this problem can not be solved using a hash table. </span></p><p> through the first traversal to find the most occurrences of the two elements (using this principle), this time can not get the exact number of occurrences, so get two occurrences of the most elements, Need to traverse again to get the number of occurrences of the accuracy of the array length of One-third </p><p></p><p> code: </p>import java.util.ArrayList; Import Java.util.list;public class Leetcode229_majorityelementii {/** * @param args */public static void main (string[] Arg s) {//TODO auto-generated method stubint[] nums={1,2,3,4,5,6,1,1,1,2,2,2,1,2}; List<integer> result=majorityelement (Nums); for (int i=0;i<result.size (); i++) System.out.println ( Result.get (i));} public static list<integer> majorityelement (int[] nums) {list<integer> result=new Arraylist<integer > (); int Majornum1=0,majornum2=0,majorcount1=0,majorcount2=0;for (int i=0;i<nums.length;i++) {if (majorcount1==0) {majorNum1=nums[i]; Majorcount1=1;} else if (Majornum1==nums[i]) {majorcount1++;} else if (majorcount2==0) {majornum2=nums[i];majorcount2=1;} else if (Majornum2==nums[i]) {majorcount2++;} else {majorcount1--;majorcount2--;}} majorcount1=0;majorcount2=0;for (int i=0;i<nums.length;i++) {if (NUMS[I]==MAJORNUM1) majorcount1++;if (nums[i]== majorNum2) majorcount2++;} if (MAJORCOUNT1&GT;NUMS.LENGTH/3) Result.add (MAJORNUM1); if (majorcount2>nums.length/3&&majornum2!=            MAJORNUM1) Result.add (majorNum2); return result; }}

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LeetCode229 majorityelementii Java