Color Classification (Lintcode)

The color class is given an array of red, white, blue, and length n, and the array elements are categorized so that the elements of the same color are adjacent and sorted in the order of red, white, and blue.

We can use integers 0,1 and 2 to represent red, white, and blue, respectively.

Sample Example Note

You cannot use the sort function in the code base to solve this problem

Description

A fairly straightforward solution is to use counting sorting to scan the algorithm 2 times.

First, the iterated algebraic group calculates the number of occurrences of 0,1,2, then overwrites the array with the number of occurrences of 0,1,2.

Can you think of an algorithm that only uses the constant-level extra space complexity and scans the array only once?

The code is a messy, rarely written note.

The idea is to swap 0 to the left and 2 to the right.

1 classSolution {2     //Total time: 14896 Ms3     4     /**5      * @paramnums:a List of integers which is 0, 1 or 26      * @return: Nothing7      */8      Public voidSortcolors (int[] nums) {9         intLow = 0;Ten         intHigh = Nums.length-1; One         //first find the leftmost non 0 and right non 2 A          while(Nums[low] = = 0 && Low < high) low++; -          while(Nums[high] = = 2 && low < high) high--; -  the          while(Nums[low] = = 2 | | nums[high] = = 0 && Low <High ) { -             //if the leftmost 0 is 2 or the rightmost 2 is 0, change it to the right or left -             if(Nums[low] = = 2) { -                 if(Nums[high] = = 0) { +nums[low++] = 0; -nums[high--] = 2; +}Else { Anums[high--] = 2; atNums[low] = 1; -                 } -}Else { -                 if(Nums[high] = = 0) { -nums[low++] = 0; -Nums[high] = 1; in                 } -             } to             //find the leftmost non 0 and right non 2 +              while(Nums[low] = = 0 && Low < high) low++; -              while(Nums[high] = = 2 && low < high) high--; the         } *          $          for(intI=low+1;i) {Panax Notoginseng             //at this time the leftmost non 0 and the right 2 are 1, with I find 0 or 2, swapped to the left or right side -             if(Nums[i] = = 0) { thenums[low++] = 0; +Nums[i] = 1; A}Else { the                 if(Nums[i] = = 2) { +nums[high--] = 2; -Nums[i] = 1; $                 } $             } -              -             //find the leftmost non 0 and right non 2 the              while(Nums[low] = = 0 && Low < high) low++; -              while(Nums[high] = = 2 && low < high) high--;Wuyi              the              while(Nums[low] = = 2 | | nums[high] = = 0 && Low <High ) { -                 //if the leftmost 0 is 2 or the rightmost 2 is 0, change it to the right or left Wu                 if(Nums[low] = = 2) { -                     if(Nums[high] = = 0) { Aboutnums[low++] = 0; $nums[high--] = 2; -}Else { -nums[high--] = 2; -Nums[low] = 1; A                     } +}Else { the                     if(Nums[high] = = 0) { -nums[low++] = 0; $Nums[high] = 1; the                     } the                 } the                 //find the leftmost non 0 and right non 2 the                  while(Nums[low] = = 0 && Low < high) low++; -                  while(Nums[high] = = 2 && low < high) high--; in             } the             //if the low>=i at this time, obviously to reset I the             if(I <= low) i = low+1; About         } the         //when the above loop ends, the classification is completed, only one nums is traversed, and the secondary space is constant the     } the}

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Color Classification (Lintcode)