[leetCode 75] Sort Colors

标签：java   leetcode   算法   排序   

题目链接：sort-colors

import java.util.Arrays;


/**
 * 
		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.
		
		Note:
		You are not suppose to use the library's sort function for this problem.
		
		click to show follow up.
		
		Follow up:
		A rather straight forward solution is a two-pass algorithm using counting sort.
		First, iterate the array counting number of 0's, 1's, and 2's, then overwrite array with total number of 0's, then 1's and followed by 2's.
		
		Could you come up with an one-pass algorithm using only constant space?
 *
 */


public class SortColors {

//	86 / 86 test cases passed.
//	Status: Accepted
//	Runtime: 194 ms
//	Submitted: 0 minutes ago

	//时间复杂度：O(n)  空间复杂度: O(1)
	//解题思路：
	//将数组A分为四段， A[0...i]都为0，	
	//				A[i+1...mid-1] 都为1;
	//				A[mid...right-1] 都为未处理的数
	//				A[right...length) 都为2;	
	
    static void sortColors(int[] A) {

    	int left = -1;
        int right = A.length;
        int mid = 0;        
        
        while(mid < right && A[mid] == 0) left = (mid++);
        
        while(mid < right && A[right - 1] == 2) right--; 
        
        while(mid < right) {        	
        	while(mid < right && A[mid] == 1) mid ++;        	
        	if(mid == right) break;        	
        	if(A[mid] == 0) exch(A, ++left, mid++);        	
        	else if(A[mid] == 2) exch(A, mid, --right);        	
        }
//        System.out.println(Arrays.toString(A));
    }
    static void exch(int[] A, int i, int j) {
    	int temp = A[i];
    	A[i] = A[j];
    	A[j] = temp;
    }
    
	public static void main(String[] args) {
		sortColors(new int[]{1});
		sortColors(new int[]{2, 0, 0});
		sortColors(new int[]{0, 1, 2, 0, 1, 1, 2, 2, 0, 1, 2});

	}

}

[leetCode 75] Sort Colors

标签：java   leetcode   算法   排序   

原文地址：http://blog.csdn.net/ever223/article/details/44491675