LeetCode高频100【2】

目录

【2】

	#	Title	Acceptance	Difficulty
	46	Permutations	53.0%	Medium
	48	Rotate Image	46.4%	Medium
	49	Group Anagrams	44.3%	Medium
	53	Maximum Subarray	42.6%	Easy
	55	Jump Game	31.1%	Medium
	56	Merge Intervals	34.6%	Medium
	62	Unique Paths	46.1%	Medium
	64	Minimum Path Sum	45.2%	Medium
	70	Climbing Stairs	43.2%	Easy
	72	Edit Distance	36.2%	Hard

46. Permutations（Medium）

Given a collection of distinct integers, return all possible permutations.
Example:
Input: [1,2,3]
Output:
[
[1,2,3],
[1,3,2],
[2,1,3],
[2,3,1],
[3,1,2],
[3,2,1]
]

# my solution, Runtime: 48 ms, faster than 93.95% of Python3 , 回溯法
class Solution:
    def permute(self, nums: List[int]) -> List[List[int]]:
        def dfs(nums,temp):
            if len(temp)==n:
                ans.append(temp)
                temp=[]
            for i in nums:
                dfs(list(set(nums)-set(temp+[i])),temp+[i])
        ans=[]
        n=len(nums)
        dfs(nums,[])
        return ans

48. Rotate Image（Medium）

You are given an n x n 2D matrix representing an image.
Rotate the image by 90 degrees (clockwise).
Note:
You have to rotate the image in-place, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the rotation.
Example:

# my solution, Runtime: 32 ms, faster than 99.57% of Python3
class Solution:
    def rotate(self, matrix: List[List[int]]) -> None:
        """
        Do not return anything, modify matrix in-place instead.
        """
        n=len(matrix)
        temp=[[ -1 for j in range(n)] for i in range(n)] #不能用[[-1]*n]*n
        for i in range(n):
            for j in range(n):
                temp[j][n-1-i]=matrix[i][j]
        matrix[:]=temp[:]
        
# Runtime: 40 ms, faster than 81.85% of Python3
class Solution(object):
    def rotate(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: void Do not return anything, modify matrix in-place instead.
        """
        n = len(matrix)
        for i in xrange(n):
            for j in xrange(n):
                if i < j:
                    matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]
        for l in matrix:
            l.reverse()

49. Group Anagrams（Medium）

Given an array of strings, group anagrams together.
Example:
Input: ["eat", "tea", "tan", "ate", "nat", "bat"],
Output:
[
["ate","eat","tea"],
["nat","tan"],
["bat"]
]
Note:
All inputs will be in lowercase.
The order of your output does not matter.

# Approach 1: Categorize by Sorted String, T(n)=O(NKlogK), S(n)=O(NK)
# Runtime: 116 ms, faster than 82.68% of Python3
class Solution:
    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
        ans = {}  # S(n)=O(NK)
        for s in strs: # O(N)
            if tuple(sorted(s)) in ans: # O(KlogK),list不能作为key，tuple可以
                ans[tuple(sorted(s))].append(s)
            else:
                ans[tuple(sorted(s))]=[s]
        return list(ans.values())

# Approach 2: Categorize by Count, T(n)=O(NK), S(n)=O(NK)
class Solution:
    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
        ans = {} # S(n)=O(NK)
        for s in strs: # O(N)
            count=[0]*26
            for c in s:
                count[ord(c)-ord('a')]+=1 # O(K)
            if tuple(count) in ans:
                ans[tuple(count)].append(s)
            else:
                ans[tuple(count)]=[s]
        return list(ans.values())

53. Maximum Subarray（Easy）

Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.
Example:
Input: [-2,1,-3,4,-1,2,1,-5,4],
Output: 6
Explanation: [4,-1,2,1] has the largest sum = 6.