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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?
题目大意：数组包括0,1,2，依照大小排列这些数字。
利用STL的sort1句话就解决了。
class Solution {
public:
void sortColors(vector<int>& nums) {
sort(nums.begin(),nums.end());
}
};算法思想很简单，把所有的0交换到队首，把所有的1交换到队尾
class Solution {
public:
void sortColors(vector<int>& nums) {
int start = 0;
int end = nums.size()-1;
for(int i = 0 ; i<nums.size();i++)//把0交换到前面
{
if(nums[i]==0) {
if(i!=start) swap(nums[i],nums[start]);
start++;
}
}
for(int i = nums.size()-1 ; i>=start ;i--)//把2交换到尾部
{
if(nums[i]==2){
if(i!=end) swap(nums[i],nums[end]);
end--;
}
}
}
};保持3个指针i、j和k，从0~i表示1，i+1~j表示1，k~nums.size()表示2
代码也比较简单：
class Solution {
public:
void sortColors(vector<int>& nums) {
int i = 0, j = i, k = nums.size() - 1;
while(j <= k){
if(nums[j] == 0) swap(nums[i++], nums[j++]);//0交换到前面
else if(nums[j] == 1) j++;//1保持不动
else swap(nums[k--], nums[j]);//2交换到尾部
}
}
};
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