Description

Word Search

Given a 2D board and a word, find if the word exists in the grid.

The word can be constructed from letters of sequentially adjacent cell, where “adjacent” cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.

Example:

board =
[
  ['A','B','C','E'],
  ['S','F','C','S'],
  ['A','D','E','E']
]

Given word = "ABCCED", return true.
Given word = "SEE", return true.
Given word = "ABCB", return false.

board中查找是否存在相连(相邻)的字符与给定的单词对应。

很像一个走迷宫的题目,搜索下一步是否与单词的字母匹配,是则走到该位置。

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Description

Subsets

Given a set of distinct integers, nums, return all possible subsets (the power set).

Note: The solution set must not contain duplicate subsets.

Example:

Input: nums = [1,2,3] Output: [ [3], [1], [2], [1,2,3], [1,3], [2,3], [1,2], [] ]

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Description

Combinations

Given two integers n and k, return all possible combinations of k numbers out of 1 … n.

Example:

Input: n = 4, k = 2
Output:
[
  [2,4],
  [3,4],
  [2,3],
  [1,2],
  [1,3],
  [1,4],
]

给出两个整数nk,输出所有包含k个数字的组合,数字范围1..n

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Description

Minimum Window Substring

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:

  • If there is no such window in S that covers all characters in T, return the empty string “”.
  • If there is such window, you are guaranteed that there will always be only one unique minimum window in S.

给出两个字符串ST,在S中找到一个最短的子串,它包含T中所有的字符。

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Description

Sort Colors

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 suppose 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]
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 a one-pass algorithm using only constant space?

对一个只包含012三个数字的数组进行排序,要求时间复杂度O(n),空间复杂度O(1)

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Description

Search a 2D Matrix

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

  • Integers in each row are sorted from left to right.
  • The first integer of each row is greater than the last integer of the previous row.

Example 1:

Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]

target = 3
Output: true

target = 13
Output: false

矩阵中有序存地放着一些数字,查找是否存在某个数字。

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Description

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]
]

Example 2:

Input:
[
  [0,1,2,0],
  [3,4,5,2],
  [1,3,1,5]
]
Output:
[
  [0,0,0,0],
  [0,4,5,0],
  [0,3,1,0]
]

Follow up:

  • A straight forward solution using O(mn) space is probably a bad idea.
  • A simple improvement uses O(m + n) space, but still not the best solution.
  • Could you devise a constant space solution?

m x n的矩阵中,如果一个元素为0,则将它所在的行和列都置为0

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Description

Edit Distance

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:

  1. Insert a character
  2. Delete a character
  3. Replace a character

Example 1:

Input: word1 = “horse”, word2 = “ros” Output: 3 Explanation: horse -> rorse (replace ‘h’ with ‘r’) rorse -> rose (remove ‘r’) rose -> ros (remove ‘e’)

>
> **Example 2:**
>
> ```
Input: word1 = "intention", word2 = "execution"
Output: 5
Explanation:
intention -> inention (remove 't')
inention -> enention (replace 'i' with 'e')
enention -> exention (replace 'n' with 'x')
exention -> exection (replace 'n' with 'c')
exection -> execution (insert 'u')

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Author's picture

SelFree

城市永远年轻,而我们终将老去

Programer

ChengDu·China
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