2014
11-19

# LeetCode-Sort Colors[排序]

### Sort Colors

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.

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?

// LeetCode, Sort Colors
// Counting Sort
// 时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:
void sortColors(int A[], int n) {
int counts[3] = { 0 }; // 记录每个颜色出现的次数

for (int i = 0; i < n; i++)
counts[A[i]]++;

for (int i = 0, index = 0; i < 3; i++)
for (int j = 0; j < counts[i]; j++)
A[index++] = i;

}
};


// LeetCode, Sort Colors
// 双指针，时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:
void sortColors(int A[], int n) {
// 一个是red的index，一个是blue的index，两边往中间走
int red = 0, blue = n - 1;

for (int i = 0; i < blue + 1;) {
if (A[i] == 0)
swap(A[i++], A[red++]);
else if (A[i] == 2)
swap(A[i], A[blue--]);
else
i++;
}
}
};


// LeetCode, Sort Colors
// use partition()
// 时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:
void sortColors(int A[], int n) {
partition(partition(A, A + n, bind1st(equal_to<int>(), 0)), A + n,
bind1st(equal_to<int>(), 1));
}
};


// LeetCode, Sort Colors
// 重新实现 partition()
// 时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:
void sortColors(int A[], int n) {
partition(partition(A, A + n, bind1st(equal_to<int>(), 0)), A + n,
bind1st(equal_to<int>(), 1));
}
private:
template<typename ForwardIterator, typename UnaryPredicate>
ForwardIterator partition(ForwardIterator first, ForwardIterator last,
UnaryPredicate pred) {
auto pos = first;

for (; first != last; ++first)
if (pred(*first))
swap(*first, *pos++);

return pos;
}
};


1. 呜呜呜呜呜呜呜呜呜呜呜呜呜呜呜呜哇哇哇哇哇哇哇哇呜呜呜呜哇哇哇哇呜呜呜呜呜呜呜呜呜呜呜呜呜呜呜呜呜呜呜呜呜呜呜呜，我可怜的朝颜啊

2. Often We don’t set up on weblogs, but I would like to condition that this established up really forced me individually to do this! considerably outstanding publish

3. 第23行：
hash = -1是否应该改成hash[s ] = -1

因为是要把从字符串s的start位到当前位在hash中重置

修改提交后能accept，但是不修改居然也能accept

4. 在方法1里面：

//遍历所有的边，计算入度
for(int i=0; i<V; i++)
{
degree = 0;