Leetcode:third Maximum Number

Given a non-empty array of integers,returnThe third maximum number in ThisArray. If It does not exist,returnThe maximum number. The time complexity must is in O (n). Example1: Input: [3, 2, 1]output:1explanation:the Third Maximum is1. Example2: Input: [1, 2]output:2explanation:the Third maximum does not exist, so the maximum (2) is returned instead. Example3: Input: [2, 2, 3, 1]output:1Explanation:note The third maximum here means the third maximum distinct number. Both numbers with value2 is both considered as second maximum.

My Solution:

1  Public classSolution {2      Public intThirdmax (int[] nums) {3         LongFstmax =Long.min_value;4         LongSndmax =Long.min_value;5         LongThirdmax =Long.min_value;6          for(Longeach:nums) {7             if(Each >Fstmax) {8                 Longtemp =Fstmax;9Fstmax =Each ;Teneach =temp; One             } A             if(Each<fstmax && each>Sndmax) { -                 Longtemp =Sndmax; -Sndmax =Each ; theeach =temp; -             } -             if(Each<fstmax && Each<sndmax && each>Thirdmax) { -Thirdmax =Each ; +             } -         } +         returnThirdmax==long.min_value? (int) Fstmax: (int) Thirdmax; A     } at}

Highest votes in discussion

1  Public intThirdmax (int[] nums) {2         intMax, Mid, small, count;3max = Mid = small =Integer.min_value;4Count = 0;//Count How many top elements has been found.5 6          for(intx:nums) {7             //Skip Loop if Max or mid elements is duplicate. The purpose is to avoiding right shift.8             if(x = = Max | | x = =mid) {9                 Continue;Ten             } One  A             if(X >max) { -                 //Right shift -small =mid; theMID =Max; -  -Max =x; -count++; +}Else if(X >mid) { -                 //Right shift +small =mid; A  atMID =x; -count++; -}Else if(x >= Small) {//if small duplicated, that ' s find, there's no shift and need to increase count. -small =x; -count++; -             } in         } -  to         //"Count" is the used for checking whether found top 3 maximum elements. +         if(Count >= 3) {  -             returnSmall; the}Else { *             returnMax; $         }Panax Notoginseng}

Leetcode:third Maximum Number