Given an array with n objects colored red, white or blue, sort them so that objects of the same color are adjacent, with the colors in the order red, white and blue.
Here, we will use the integers 0, 1, and 2 to represent the color red, white, and blue respectively.
Solution #2: (one-pass solution)
https://linchicoding.wordpress.com/2014/08/26/leetcode-sort-colors/
Solution #1: (two-pass counting sort as the follow up says)
http://group.jobbole.com/8144/
http://yucoding.blogspot.com/2013/05/leetcode-questions-99-sort-colors.html
Read full article from 一天一学: LeetCode - Sort Colors
Here, we will use the integers 0, 1, and 2 to represent the color red, white, and blue respectively.
Solution #2: (one-pass solution)
https://linchicoding.wordpress.com/2014/08/26/leetcode-sort-colors/
public void sortColors(int[] A) { if(A.length==0)return; int red = 0; int blue = A.length-1; int i = 0; while(i <= blue){ if(A[i]==0){ //swap with red A[i] = A[red]; A[red] = 0; red++; } if(A[i]==2){ //swap with blue A[i] = A[blue]; A[blue] = 2; blue--; i--; //---!!!--important--for case A[i]=0 then swaped to 2, it go through both branches //--but for case A[i] = 2 then switch to 0, need to decrease i so can enter first branch } i++; } }public void sortColors(int[] A) { if(A == null || A.length == 0 || A.length == 1) return; // one-pass solution int red = 0, blue = A.length - 1, tmp, i = 0; // stop looping when current >= blue while(i <= blue) { // if color is red, move to the front if(A[i] == 0) { // when cur > red, switch if(i > red) { tmp = A[red]; A[red] = A[i]; A[i] = tmp; red++; } // when cur <= red, no need to switch, just move both to next else { i++; red++; } } // if color is blue, move to the end else if(A[i] == 2) { // when cur < blue, switch if(i < blue) { tmp = A[blue]; A[blue] = A[i]; A[i] = tmp; blue--; } // when cur >= blue, end the loop else { return; } } // if color is white, skip else { i++; } } }public void sortColors(int[] A) { if(A == null || A.length == 0 || A.length == 1) return; int red = 0, white = 0, blue = 0; for(int i = 0; i < A.length; i++) { if(A[i] == 0) red++; else if(A[i] == 1) white++; else blue++; } for(int i = 0; i < red; i++) A[i] = 0; for(int i = red; i < red + white; i++) A[i] = 1; for(int i = red + white; i < A.length; i++) A[i] = 2; }http://yucoding.blogspot.com/2013/05/leetcode-questions-99-sort-colors.html
Read full article from 一天一学: LeetCode - Sort Colors
