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学习笔记

# Title Solutions
-1 sort 冒泡/选择/插入/归并/快速/归并排序(Go,Py)
191 number-of-1-bits 依次除2(Go,Py),按位与n-1(Go,Py)
190 reverse_bits 迭代(Go,Py),分治位运算(Go,Py)
338 counting-bits 动态规划1(Go,Py),动态规划2(Go,Py)
231 power-of-two 位运算1(Go,Py),位运算2(Go,Py)
326 power-of-three 累除法(Go,Py),累乘法(Go,Py)
342 power-of-four 累除法(Go,Py),位运算(Go,Py)
146 lru-cache 双向链表+字典(Go,Py)
1122 relative-sort-array 计数排序(Go,Py)
56 merge-intervals 计数排序(Go,Py)
453 reverse-pairs 二分查找和插入(Go,Py)

题解

191. number-of-1-bits

  1. 除2:依次除2,看余数是否是1
  2. 按位与n-1: n & n-1 会消去最后一个1

190. reverse-bits

  1. 除2:
  2. 分治:依次以16、8、4、2、1两两交换

338. counting-bits

  1. 动态规划1: dp[i] = dp[i//2] + i % 2
  2. 动态规划2: dp[i] = dp[i & (i-1)] + 1

231. power-of-two

  1. 位运算1: n & (n-1)为消去最后一个1
  2. 位运算2: n & -n会保留最后一个1,其他位都置0

326. power-of-three

  1. 累除法
  2. 累乘法

342. power-of-four

  1. 累除法
  2. 位运算

146. lru_cache

  1. 双向链表+字典

493. reverse-pairs

  1. 二分查找: 逆序遍历nums, num[i]在有序数组arr的位置即为nums[i]的对数,随后将num[i]*2二分查找插入arr