leetCode 75.Sort Colors (颜色排序) 解题思路和方法

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?

思路：如果空间上不做要求，这题还是比较简单的。可以再声明一个m,n矩阵，然后搜索原矩阵，是0，置于首位，是2置于末尾，中间置为1.

代码如下：

public class Solution {
    public void sortColors(int[] nums) {
        
        int[] a = new int[nums.length];
        a = Arrays.copyOf(nums, nums.length);
        
        int i = 0;
        int j = nums.length - 1;
        for(int k = 0; k < nums.length; k++){
            if(a[k] == 0){
                nums[i++] = a[k];
            }else if(a[k] == 2){
                nums[j--] = a[k];
            }
        }
        while(i <= j){
            nums[i++] = 1;
        }
    }
}

public class Solution {
    public void sortColors(int[] nums) {
        int i = -1,j = -1,k = -1;
        for(int m = 0; m < nums.length; m++){
            if(nums[m] == 0){
                nums[++k] = 2;
                nums[++j] = 1;
                nums[++i] = 0;
            }else if(nums[m] == 1){
                nums[++k] = 2;
                nums[++j] = 1;
            }else{
                nums[++k] = 2;
            }
        }
    }
}