【LeetCode with Python】 Sort Colors
2014-09-21 17:51
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原题页面:https://oj.leetcode.com/problems/sort-colors/
题目类型:两指针问题,下标计算
难度评价:★★★
本文地址:/article/1377466.html
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.
click to show follow up.
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?
定义两个两头的下标start和end,i从左至右遍历数组,发现0就和start交换(同时start-=1),发现2就和end交换(同时end-=1)。这样就能确保下标<start的元素都是0,而下标>end的元素都是2。
原题页面:https://oj.leetcode.com/problems/sort-colors/
题目类型:两指针问题,下标计算
难度评价:★★★
本文地址:/article/1377466.html
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.
click to show follow up.
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?
定义两个两头的下标start和end,i从左至右遍历数组,发现0就和start交换(同时start-=1),发现2就和end交换(同时end-=1)。这样就能确保下标<start的元素都是0,而下标>end的元素都是2。
class Solution: # @param A a list of integers # @return nothing, sort in place def sortColors(self, A): len_A = len(A) if 1 == len(A): return left = 0 right = len_A - 1 i = 0 while i <= right: if 0 == A[i]: if left == i: i += 1 else: A[left], A[i] = A[i], A[left] left += 1 elif 1 == A[i]: i += 1 else: if right == i: i += 1 else: A[right], A[i] = A[i], A[right] right -= 1
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