Leetcode75. Sort Colors

Leetcode75. Sort Colors

Given an array with n objects colored red, white or blue, sort them in-place 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.

Example:

Input: [2,0,2,1,1,0]
Output: [0,0,1,1,2,2]

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 a one-pass algorithm using only constant space?

解法

结合三路快排partition思路的应用，设定两个索引，一个从左往右滑动zero，一个从右往左滑动two，遍历nums，当nums[i]的值为1时，i++；当nums[i]的值为2时，two的值先减1，而后交换nums[i]与nums[two]，此时在观察nums[i]的值；当nums[i]的值为0时，zero++，而后交换nums[i]与nums[zero]，i++;当 i = two时，结束循环。

// 三路快速排序的思想
// 对整个数组只遍历了一遍
// 时间复杂度: O(n)
// 空间复杂度: O(1)
class Solution {
public:
    void sortColors(vector<int> &nums) {

        int zero = -1;          // [0...zero] == 0
        int two = nums.size();  // [two...n-1] == 2
        for(int i = 0 ; i < two ; ){ //注意循环没有i++
            if(nums[i] == 1){
                 i ++;
            }else if (nums[i] == 2){
                 two--;
                 swap( nums[i] , nums[two]);
            }else{ // nums[i] == 0
                 zero++;
                 swap(nums[zero] , nums[i]);
                 i++;
            }
        }
    }
};

示例 [2 0 2 0 1 0]
初始化：zero = -1, two = 6, i = 0.
① i = 0, nums[0] == 2, 则two = 5, swap(nums[0] , nums[5])，原数组变为[0 0 2 0 1 2]（注意i并没有加1）
② i = 0, nums[0] == 0, 则zero = 0, swap(nums[0] , nums[0])，i++
③ i = 1, nums[1] == 0, 则zero = 1, swap(nums[1] , nums[1])，i++
④ i = 2, nums[2] == 2, 则two = 4, swap(nums[2], nums[4])，原数组变为[0 0 1 0 2 2]
⑤ i = 2, nums[2] == 1, i++
⑥ i = 3, nums[3] == 0, zero = 2, swap(nums[2] , nums[3])，原数组变为[0 0 0 1 2 2]，i++
由于i == two，退出循环即可。

Four different solutions

// two pass O(m+n) space
void sortColors(int A[], int n) {
    int num0 = 0, num1 = 0, num2 = 0;
    
    for(int i = 0; i < n; i++) {
        if (A[i] == 0) ++num0;
        else if (A[i] == 1) ++num1;
        else if (A[i] == 2) ++num2;
    }
    
    for(int i = 0; i < num0; ++i) A[i] = 0;
    for(int i = 0; i < num1; ++i) A[num0+i] = 1;
    for(int i = 0; i < num2; ++i) A[num0+num1+i] = 2;
}

// one pass in place solution
void sortColors(int A[], int n) {
    int n0 = -1, n1 = -1, n2 = -1;
    for (int i = 0; i < n; ++i) {
        if (A[i] == 0) 
        {
            A[++n2] = 2; A[++n1] = 1; A[++n0] = 0;
        }
        else if (A[i] == 1) 
        {
            A[++n2] = 2; A[++n1] = 1;
        }
        else if (A[i] == 2) 
        {
            A[++n2] = 2;
        }
    }
}

// one pass in place solution
void sortColors(int A[], int n) {
    int j = 0, k = n - 1;
    for (int i = 0; i <= k; ++i){
        if (A[i] == 0 && i != j)
            swap(A[i--], A[j++]);
        else if (A[i] == 2 && i != k)
            swap(A[i--], A[k--]);
    }
}

// one pass in place solution
void sortColors(int A[], int n) {
    int j = 0, k = n-1;
    for (int i=0; i <= k; i++) {
        if (A[i] == 0)
            swap(A[i], A[j++]);
        else if (A[i] == 2)
            swap(A[i--], A[k--]);
    }
}