075. Sort Colors

Method One: OK, the ugly way, but write it yourself ... Draw on the thought of the Fast platoon partition

1 classSolution {2  Public:3     voidSortcolors (vector<int>&nums) {4         if(nums.size () = =0)return;5vector<int> Start (3, -1);6start[nums[0]] =0;7          for(inti =1; I < nums.size (); ++i) {8             if(Nums[i] = = Nums[i-1])Continue;9             Else {Ten                 if(Start[nums[i]] = =-1) { One                     if(Nums[i] = =2) { Astart[2] = i;Continue; -                     } -                     Else if(Nums[i] = =1) { the                         if(start[2] == -1) { -start[1] = i;Continue; -                         } -                         Else { +start[1] = start[2]; -Nums[i] =2; +nums[start[1]] =1; A++start[2]; at                         } -                     } -                     Else { -                         //cannot appear start[1] = start[2] = 1, the first element has been determined by a -                         //here start[1] and start[2] at least one is not-1 -                         if(start[2] != -1) { innums[start[2]] =0; -start[0] = start[2]; to++start[2]; +Nums[i] =2; -                             if(start[1] != -1) { thenums[start[1]] =0; *nums[start[0]] =1; $start[0] = start[1];Panax Notoginseng++start[1]; -                             } the                             Else Continue; +                         } A                         Else { thenums[start[1]] =0; +Nums[i] =1; -start[0] = start[1]; $++start[1]; $                         } -                     } -                 } the                 Else { -                     if(Nums[i] = = Nums[i-1])Continue;Wuyi                     Else { the                         if(Nums[i] = =0) { -                             if(start[2] == -1) { WuNums[i] =1; -nums[start[1]] =0; About++start[1]; $                             } -                             Else { -nums[start[2]] =0; -Nums[i] =2; A++start[2]; +                                 if(start[1] != -1) { thenums[start[1]] =0; -nums[start[2] -1] =1; $++start[1]; the                                 } the                             } the                         } the                         Else { -                             //Nums[i] = = 1, cannot be 2, and the previous one must be 2 inNums[i] =2; thenums[start[2]] =1; the++start[2]; About                         } the                     } the                 } the             } +         } -     } the};

Method Two: Two pointers, two to the middle walk, method one just start from the end, resulting in the program is too complex

1 classSolution {2      Public:3         voidSortcolors (vector<int>&A) {4             intRed =0, blue = A.size ()-1;5              for(inti =0; I < Blue +1; ) {6                 if(A[i] = =0) Swap (a[i + +), a[red + +]);7                 Else if(A[i] = =2) Swap (a[i + +), A[blue--]);8                 Else++i;9             }Ten         } One};

Method Three:

075. Sort Colors