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--]);
}
}