《数据结构》学习笔记

1.算法分析的两个主要任务:正确性(不变性 + 单调性) + 复杂度。

2.复杂度分析的主要方法:

  • 迭代:级数求和;
  • 递归:递归跟踪 + 递推方程
  • 猜测 + 验证

3.级数:

(1)算术级数:与末项平方同阶
T ( n ) = 1 + 2 + ⋯ + n = n ( n + 1 ) 2 = O ( n 2 ) T(n) = 1 + 2 + \cdots + n = \frac{n(n+1)}{2} = O(n^2) \notag T(n)=1+2++n=2n(n+1)=O(n2)
(2)幂方级数:比幂次高出一阶
∑ k = 0 n k d ≈ ∫ 0 n x d + 1 d x = 1 d + 1 x d + 1 ∣ 0 n = 1 d + 1 n d + 1 = O ( n d + 1 ) \sum \limits _{k=0} ^ n k^d \approx \int _0 ^n x^{d+1} \text{d} x = \frac{1}{d+1} x^{d+1} \bigg|_0^n = \frac{1}{d+1} n^{d+1} = O(n^{d+1}) \notag k=0nkd0nxd+1dx=d+11xd+1 0n=d+11nd+1=O(nd+1)
例如:
T 2 ( n ) = 1 2 + 2 2 + ⋯ + n 2 = n ( n + 1 ) ( 2 n + 1 ) 6 = O ( n 3 ) T 3 ( n ) = 1 3 + 2 3 + ⋯ + n 3 = n 2 ( n + 1 ) 2 4 = O ( n 4 ) T 4 ( n ) = 1 4 + 2 4 + ⋯ + n 4 = n ( n + 1 ) ( 2 n + 1 ) ( 3 n 2 + 3 n − 1 ) 30 = O ( n 5 ) \begin{aligned} & T_2(n) = 1^2 + 2 ^ 2 +\cdots + n^2 = \frac{n(n+1)(2n+1)}{6} = O(n^3) \\ & T_3(n) = 1^3 + 2 ^3 +\cdots + n^3 = \frac{n^2(n+1)^2}{4} = O(n^4) \\ & T_4(n) = 1^4 + 2^4 +\cdots + n^4 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30} = O(n^5) \end{aligned}\notag T2(n)=12+22++n2=6n(n+1)(2n+1)=O(n3)T3(n)=13+23++n3=4n2(n+1)2=O(n4)T4(n)=14+24++n4=30n(n+1)(2n+1)(3n2+3n1)=O(n5)
(3)几何级数:与末项同阶
T a ( n ) = a 0 + a 1 + ⋯ + a n = a n + 1 − 1 a − 1 = O ( a n ) T_a(n) = a^0 + a^1+\cdots+a^n = \frac{a^{n+1}-1}{a-1} = O(a^n) \notag Ta(n)=a0+a1++an=a1an+11=O(an)
例如:
1 + 2 + 4 + ⋯ + 2 n = 2 n + 1 − 1 = O ( 2 n + 1 ) = O ( 2 n ) 1+2+4+\cdots+2^n = 2^{n+1}-1=O(2^{n+1})=O(2^n)\notag 1+2+4++2n=2n+11=O(2n+1)=O(2n)
(4)收敛级数:
1 1 × 2 + 1 2 × 3 + 1 3 × 4 + ⋯ + 1 ( n − 1 ) × n = 1 − 1 n = O ( 1 ) 1 + 1 2 2 + ⋯ + 1 n 2 < 1 + 1 2 2 + ⋯ + = π 2 6 = O ( 1 ) 1 3 + 1 7 + 1 8 + 1 15 + 1 24 + 1 26 + 1 31 + 1 35 + ⋯ = 1 = O ( 1 ) \begin{aligned} & \frac{1}{1\times 2} + \frac{1}{2 \times 3} + \frac{1}{3\times4} +\cdots +\frac{1}{(n-1)\times n} = 1-\frac 1 n = O(1) \\ & 1 + \frac{1}{2^2} + \cdots + \frac{1}{n^2} < 1 + \frac{1}{2^2} + \cdots + = \frac{\pi ^2}{6} = O(1) \\ & \frac{1}{3} + \frac{1}{7} + \frac{1}{8} + \frac{1}{15} + \frac{1}{24} + \frac{1}{26} + \frac{1}{31} + \frac{1}{35} + \cdots = 1 = O(1) \end{aligned} \notag 1×21+2×31+3×41++(n1)×n1=1n1=O(1)1+221++n21<1+221++=6π2=O(1)31+71+81+151+241+261+311+351+=1=O(1)
(5)两个重要级数:

  • 调和级数:

h ( n ) = 1 + 1 2 + 1 3 + ⋯ + 1 n = Θ ( log ⁡ n ) \begin{aligned} & h(n) = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} = \Theta(\log n) \end{aligned} \notag h(n)=1+21+31++n1=Θ(logn)

  • 对数级数:

log ⁡ 1 + log ⁡ 2 + log ⁡ 3 + ⋯ + log ⁡ n = log ⁡ ( n ! ) = Θ ( n log ⁡ n ) \log 1 + \log 2 +\log 3 + \cdots + \log n = \log (n!) = \Theta(n \log n) \notag log1+log2+log3++logn=log(n!)=Θ(nlogn)

4.一些循环的复杂度估计:

(1)循环体如下:

for (int i = 1; i < n; i <<= 1)
    for (int j = 0; j < i; j++)
        O1_Operation(i, j);

其时间复杂度的计算如下:
几何级数: 1 + 2 + 4 + ⋯ + 2 ⌊ log ⁡ 2 ( n − 1 ) ⌋ = ∑ k = 0 ⌊ log ⁡ 2 ( n − 1 ) ⌋ 2 k ( let  k = log ⁡ 2 i ) = 2 ⌈ log ⁡ 2 n ⌉ − 1 = O ( n ) \begin{aligned} \text{几何级数:} \\ & 1 + 2 + 4 + \cdots + 2^{\lfloor \log_2 (n-1) \rfloor} \\ = &\sum \limits _{k=0} ^ {\lfloor \log_2 (n-1) \rfloor} 2^k \quad (\text{let}\ k = \log_2 i) \\ = & 2^{\lceil \log _2 n \rceil} - 1 \\ = & O(n) \end{aligned} \notag 几何级数:===1+2+4++2log2(n1)⌋k=0log2(n1)⌋2k(let k=log2i)2log2n1O(n)
(2)前置知识:数列 { n ⋅ 2 n } \{n\cdot 2^n\} {n2n} 的前 n n n 项和: S n = ( n − 1 ) ⋅ 2 n + 1 + 2 S_n = (n-1)\cdot 2^{n+1}+2 Sn=(n1)2n+1+2

循环体如下:

for (int i = 0; i <= n; i++)
    for (int j = 1; j < i; j += j)
        O1_Operation(i, j);

时间复杂度的计算如下:
几何级数: ∑ k = 0 n ⌈ log ⁡ 2 i ⌉ = O ( n log ⁡ n ) ( i = 0 , 1 , 2 , 3 ∼ 4 , 5 ∼ 8 , 9 ∼ 16 , ⋯   ) =   0 + 0 + 1 + 2 × 2 + 3 × 2 2 + 4 × 2 4 + ⋯ =   ∑ k = 0 ⋯ log ⁡ n ( k × 2 k − 1 ) =   O ( log ⁡ n × 2 log ⁡ n ) \begin{aligned} \text{几何级数:} & \sum \limits _{k=0}^n {\lceil \log _2 i \rceil} = O(n \log n) \\ & (i = 0, 1, 2, 3\sim4, 5\sim 8, 9 \sim 16, \cdots) \\ = &\ 0 + 0 + 1 + 2\times 2 + 3 \times 2^2 + 4 \times 2^4 +\cdots \\ = & \ \sum _{k=0\cdots \log n} (k \times 2^{k-1}) \\ = & \ O(\log n \times 2^{\log n}) \end{aligned} \notag 几何级数:===k=0nlog2i=O(nlogn)(i=0,1,2,34,58,916,) 0+0+1+2×2+3×22+4×24+ k=0logn(k×2k1) O(logn×2logn)
5.算法正确性的证明:(以起泡排序为例)

  • 不变性:经 k k k 轮扫描交换后,最大的 k k k 个元素必然就位;
  • 单调性:经 k k k 轮扫描交换后,问题规模缩减至 n − k n-k nk
  • 正确性:经至多 n n n 趟扫描后,算法必然终止,且能给出正确解答。

6.封底估算(Back-Of-The-Envelope Calculation):

(1)1 day = 24h x 60min x 60sec ≈ \approx 25 x 4000 = 10^5 sec

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