leetcode 75. Sort Colors-颜色排序|双指针

原题链接：75. Sort Colors
【思路1-Python】-双指针|T=O(n)|M=O(1)

申请两枚指针，head 和 tail，用 i 进行遍历，当 num[i] == 0时，交换当前位置和头指针处值，当 nums[i] == 2时，交换当前位置和尾指针处值，当 nums[i] == 1时，不进行交换：

class Solution(object):
def sortColors(self, nums):
"""
:type nums: List[int]
:rtype: void Do not return anything, modify nums in-place instead.
"""
head, tail, i = 0, len(nums)-1, 0
while i <= tail :
if nums[i] == 0 :
nums[i], nums[head] = nums[head], nums[i]
head += 1
i += 1
elif nums[i] == 1 : i += 1
else :
nums[i], nums[tail] = nums[tail], nums[i]
tail -= 1

86 / 86 test
cases passed. Runtime: 52
ms  Your runtime beats 46.69% of pythonsubmissions.

【思路2-Java】-双指针|T=O(n)|M=O(n)

申请一个数组，初始化数组值为1，也是用双指针。当 num[i] == 0时，覆盖头指针处值，当 nums[i] == 2时，覆盖尾指针处值，当 nums[i] == 1时，不覆盖，最后将新数组的值覆盖旧数组：

public class Solution {
public static void sortColors(int[] nums) {
int len = nums.length;
int[] tmp = new int[len];
for(int i = 0; i < len; i++)      //初始化数组为1
tmp[i] = 1;
int first = 0, last = len - 1;
for(int i=0; i<len; i++)          //遍历一次
if(nums[i] == 0) tmp[first++] = 0;
else if(nums[i] == 2) tmp[last--] = 2;
for(int i=0; i<len; i++) nums[i] = tmp[i];
}
}

86 / 86 test
cases passed. Runtime: 1
ms  Your runtime beats 4.38% of javasubmissions.