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//算法是一个程序员的基础，也是重中之重，我希望能重头系统的学习一遍算法。
Analysis of Algorithm
在程序领域，what‘s more important than the perfermance?
正确性，简洁性，健壮性。features，模块化，security, user friendly。
如果这些都比性能重要，那为什么我们还要学习性能和算法呢？
性能在很多时候，是用户体验的保证，
1，性能的好与坏，可能决定程序的可行于不可行。
2，算法是一种描述程序行为的语言。
3，是用户体验和安全的保障。
4，性能相当于一种货币，是一种标准和基础。
5，算法是一种刺激的东西。
startup with a simple problem
Sorting:
Input Array <a1, a2, ...,an> of number s.
Output permutation <a1‘, a2‘, ..., an‘> a1‘ <= a2’ <= ...<= an;
Insertion sort:
1 void InsertionSort(vector<int>& nums, int n) 2 { 3 int key; 4 //插入排序 5 //每一个pass，在有序数列0 - i-1 中插入第i个元素 6 for (int i = 1; i < n; ++i) 7 { 8 key = nums[i]; 9 for (int j = i - 1; j >= 0; --j) 10 { 11 if (nums[j] >= key) 12 { 13 //比key大的向前挪一位 14 nums[j + 1] = nums[j]; 15 if (j == 0) nums[j] = key; 16 } 17 else 18 { 19 nums[j + 1] = key; 20 break; 21 } 22 23 24 } 25 } 26 }
issue of the running time:
1,depend on the input itself.(e.g. already sorted)
2,depend on the input size.(6 elem VS. 6^9 elem)
3,Want upper bounds.(guarantee to the users)
Kinds of analysis
Worst case:(Usually)
T(n) = max time on any input of size n;
Average case:(Sometimes)
T(n) = expected time over all inputs of size n;
Best case:(bogus)
What is the sort‘s w-c time:
Depends on your computer;
relative speed (on same machine)
absolute speed (on diff machine)
Big Idea:渐进分析
1,Ignore the machine dependent
2,look at growth of T(n) n -> 无穷
θ ：弃去低阶项，并忽略前面的常熟因子
当n趋近于无穷大的时候，性能只受到最高项的影响。(可以做到忽略不同计算机上性能的影响，消除特异性的麻烦)
Insertion sort :T(n) = ∑(from 2 - n) j = 1 + 2 + 3 + 4 ... = θ(n^2)
is insertion sort fast?
Merge sorting: A[1, 2, .. , n]
1 if n = 1, done;-----------------------------------------------------θ(1)
2 Recursively sort A[1, 2, .. , [n/2]] and A[[n/2]+1, .., n];-----2θ(n/2)
3 Merge sorted list--------------------------------------------------θ(n)
递归树
the height of the tree is lg(n) the work of each level is Cn, So the total work is Theta(nlgn)
void Merge(vector<int>& nums, int begin, int half, int end) { int n1, n2; int *left = NULL, *right = NULL; n1 = half - begin + 1; n2 = end - half; left = (int *)malloc(sizeof(int) * n1); right = (int *)malloc(sizeof(int) * n2); for (int i = 0; i < n1; i++) { left[i] = nums[begin + i]; } for (int j = 0; j < n2; j++) { right[j] = nums[half + 1 + j]; } int k = begin; int i = 0, j = 0; while (i < n1 && j < n2) { if (left[i] > right[j]) { nums[k++] = right[j++]; } else { nums[k++] = left[i++]; } } for (; i < n1; i++) { nums[k++] = left[i]; } for (; j < n2; j++) { nums[k++] = right[j]; } } void MergeSort(vector<int>& nums, int begin, int end) { int half = (end - begin) / 2; if (begin < end){ MergeSort(nums, begin , begin + half); MergeSort(nums, begin + half + 1, end); Merge(nums, begin, begin + half, end); } }
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原文地址：http://www.cnblogs.com/stormhan/p/5323610.html
