lintcode-medium-Sort Colors II
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Given an array of n objects with k different colors (numbered from 1 to k), sort them so that objects of the same color are adjacent, with the colors in the order 1, 2, ... k.
Example
Given colors=[3, 2, 2, 1, 4]
, k=4
, your code should sort colors in-place to[1, 2, 2, 3, 4]
.
Challenge
A rather straight forward solution is a two-pass algorithm using counting sort. That will cost O(k) extra memory. Can you do it without using extra memory?
class Solution { /** * @param colors: A list of integer * @param k: An integer * @return: nothing */ public void sortColors2(int[] colors, int k) { // write your code here if(colors == null || colors.length == 0) return; int start = 0; int end = colors.length - 1; int count = 0; while(count < k){ int left = start; int right = end; int max = Integer.MIN_VALUE; int min = Integer.MAX_VALUE; for(int i = start; i <= end; i++){ max = Math.max(max, colors[i]); min = Math.min(min, colors[i]); } while(left < right && colors[left] == min) left++; while(left < right && colors[right] == max) right--; int i = left; while(i <= right){ if(colors[i] == min){ swap(colors, i, left); left++; while(left < right && colors[left] == min) left++; i = left; } else if(colors[i] == max){ swap(colors, i, right); right--; while(left < right && colors[right] == max) right--; } else{ i++; } } start = left; end = right; count += 2; } return; } public void swap(int[] nums, int i, int j){ int temp = nums[i]; nums[i] = nums[j]; nums[j] = temp; return; } }
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