标签：+= use replace ber its tin val already append
LeetCode66. Plus One - Easy
LeetCode67. Add Binary - Easy
LeetCode69. Sqrt(x) - Easy 二分找
LeetCode70. Climbing Stairs - Easy dp
LeetCode71. Simplify Path - Medium
Given an absolute path for a file (Unix-style), simplify it. Or in other words, convert it to the canonical path.
In a UNIX-style file system, a period . refers to the current directory. Furthermore, a double period .. moves the directory up a level. For more information, see: Absolute path vs relative path in Linux/Unix
Note that the returned canonical path must always begin with a slash /, and there must be only a single slash / between two directory names. The last directory name (if it exists) must not end with a trailing /. Also, the canonical path must be the shortest string representing the absolute path.
Example 1:
Input: "/home/" Output: "/home" Explanation: Note that there is no trailing slash after the last directory name.
思路：注意‘/..hidden/‘这一类例子。
用‘/‘split之后，用stack记录碰到的内容，且判断是否是‘.‘或者是‘..‘。
class Solution: def simplifyPath(self, path: str) -> str: if not path or len(path) == 0: return " " res = ‘‘ stack = [] i = 0 path = path.split(‘/‘) print(path) for i in range(len(path)): if not path[i] or path[i] == ‘.‘: continue elif path[i] == ‘..‘: if stack: stack.pop() else: continue else: stack.append(path[i]) for i in range(len(stack)): res += ‘/‘+stack[i] if not res: res = ‘/‘ return res
LeetCode72. Edit Distance - Hard
Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.
You have the following 3 operations permitted on a word:
Example 1:
Input: word1 = "horse", word2 = "ros" Output: 3 Explanation: horse -> rorse (replace ‘h‘ with ‘r‘) rorse -> rose (remove ‘r‘) rose -> ros (remove ‘e‘)
思路：和text regex一样的思路，用一个len(s1)+1 * len(s2) +1 的dp来记录状态。
class Solution: def minDistance(self, word1: str, word2: str) -> int: if not word1 or len(word1) == 0: return len(word2) if not word2 or len(word2) == 0: return len(word1) # the max will be len(word1) m = len(word1) n = len(word2) dp = [[0 for _ in range(n+1)] for _ in range(m+1)] temp = 1 for i in range(1,m+1): dp[i][0] = temp temp += 1 dp[0] = [i for i in range(n+1)] for i in range(1, m+1): for j in range(1, n+1): if word1[i-1] == word2[j-1]: dp[i][j] = dp[i-1][j-1] else: dp[i][j] = min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]) + 1 # print(dp) return dp[-1][-1]
LeetCode73. Set Matrix Zeroes
Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in-place.
Example 1:
Input: [ [1,1,1], [1,0,1], [1,1,1] ] Output: [ [1,0,1], [0,0,0], [1,0,1] ]
思路：用‘#’标记本来是1但是后来改为0的元素。
class Solution: def setZeroes(self, matrix: List[List[int]]) -> None: """ Do not return anything, modify matrix in-place instead. """ if not matrix or len(matrix) == 0: return [] rows = len(matrix) cols = len(matrix[0]) for i in range(rows): for j in range(cols): if matrix[i][j] == 0: row = i col = j for ii in range(rows): if matrix[ii][col] != 0: matrix[ii][col] = ‘#‘ for jj in range(cols): if matrix[row][jj] != 0: matrix[row][jj] = ‘#‘ for i in range(rows): for j in range(cols): if matrix[i][j] == ‘#‘: matrix[i][j] = 0
LeetCode74.Search a 2D Matrix -Medium
Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
Example 1:
Input: matrix = [ [1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50] ] target = 3 Output: true
Example 2:
Input: matrix = [ [1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50] ] target = 13 Output: false
思路：先对每一行的l和r进行判断target是否在这一行里，如果不在的话，再用二分来找target。注意在这一行的时候，也需要对l和r来进行判断，
以防出现这一行只有1个或者2个元素。
！！注意[[]] 需要len(matrix[0])== 0来判断
class Solution: def searchMatrix(self, matrix: List[List[int]], target: int) -> bool: if not matrix or len(matrix)==0 or len(matrix[0])== 0: return False for i in range(len(matrix)): l = 0 r = len(matrix[0])-1 if matrix[i][l] > target or matrix[i][r] < target: print(‘!!‘) continue while l < r: mid = l + (r-l) // 2 if matrix[i][mid] == target: return True elif matrix[i][mid] > target: r = mid else: l = mid + 1 if matrix[i][l] == target: return True if matrix[i][r] == target: return True return False
LeetCode75. Sort Colors - Medium
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 supposed 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]
思路：三个指针l, r, i. l指的是最后一个的0后面一位，i指的是1，r指的是第一个2开始的地方。
1 while i<=r 循环
2 如果i指到的元素是0，那么i和l指到的元素进行交换，且彼此都加1
3 如果i指到的元素是2，那么i和r指到的元素进行交换。r减1，注意的是i不能加1，因为我们不知道换来的元素是什么。
4 如果i指到的元素是1，那么只需要i加1即可。
class Solution: def sortColors(self, nums: List[int]) -> None: """ Do not return anything, modify nums in-place instead. """ if not nums or len(nums) == 0: return [] l = 0 r = len(nums)-1 i = 0 while i <= r: if nums[i] == 0: nums[l], nums[i] = nums[i], nums[l] l += 1 i += 1 elif nums[i] == 2: nums[r], nums[i] = nums[i], nums[r] r -= 1 else: i += 1
LeetCode76. Minimum Window Substring - Hard
Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).
Example:
Input: S = "ADOBECODEBANC", T = "ABC" Output: "BANC"
Note:
"".思路：two pointers + sliding window
步骤：
1 construct a counter for t and a missing for # of the characters for t
2 initialize a start, end and one of the pointers i to be zero
3 use enumerate to visit all the elements in the s. Be careful to set the index to start from 1.
1) for every element if the value in the counter for that element is greater than zero, which means t has this element, we will do missing -= 1
2) minus 1 of the value for every element in the counter
3) if the missing value is zero, it means we have already got all the characters in the t. We use i to do sliding window.
class Solution: def minWindow(self, s: str, t: str) -> str: if not s or len(s) == 0: return "" need = collections.Counter(t) missing = len(t) start = end = 0
i = 0 for j, char in enumerate(s, 1): if need[char] > 0: missing -= 1 need[char] -= 1 if missing == 0: while i < j and need[s[i]] < 0: need[s[i]] += 1 i += 1 need[s[i]] += 1 missing += 1 if end == 0 or j-i < end - start: start, end = i, j i += 1 return s[start:end]
标签：+= use replace ber its tin val already append
原文地址：https://www.cnblogs.com/sky37/p/12144642.html
