leetcode 101: Sort Colors

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本文介绍了一种用于将红、白、蓝三种颜色的数组进行排序的算法，通过两种方法实现：两遍计数排序和一遍单次扫描算法。两遍算法首先统计每种颜色的数量，然后重写数组；一遍算法则利用常数空间复杂度，在一次扫描中完成排序。

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Sort ColorsApr 9 '12

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.

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?

two pass appoarch:

public class Solution {
    public void sortColors(int[] a) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int[] c = new int[3];

    	for (int i = 0; i < a.length; i++) {
			c[a[i]] += 1;
		}
		
		for (int i = 1; i < c.length; i++) {
			c[i] += c[i - 1];
		}
		
		int[] b = Arrays.copyOf(a, a.length);

		for (int i = b.length - 1; i >=0; i--) {
			int v = b[i];
			a[c[v] - 1] = v;
			c[v]--;
		}
    }
}

one pass approach: (only suitable for 3 possible values situation. because we know the start of 0 and end of 2, we don't know the start and end of 1's)

public class Solution {
    public void sortColors(int[] a) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int l=0, h=a.length-1;
        
        for(int i=0; i<=h; ) {
            if(a[i]==0) {
                a[l++]=0;
                i++;
            } else if(a[i]==2){
                a[i] = a[h];
                a[h--] = 2;
            } else {
                i++;
            }
        }
        
        for(int i=l; i<=h; i++) {
            a[i] = 1;
        }
    }
}

class Solution {
public:
    void sortColors(int A[], int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(n<2) return;
        
        #define NUM_COLOR  3
        const int NUM = 3;
        
        vector<int> C(NUM,0);
        
        for(int i=0; i<n; i++) {
            ++C[ A[i] ];
        }
        
        for(int i=1; i<NUM_COLOR; i++) {
            C[i] += C[i-1];
        }
        
        
        int B[n];
        memcpy(B, A, n*sizeof(int));
        
        for(int i=n-1; i>=0; i--) {
            int x = B[i];
            A[ C[x] -1] = x;
            --C[x];
        }
    }
};