1.如何在排序数组中，找出给定数字出现次数？ 比如：{0,1,2,3,3,3,3,3,3,3,3,4,5,6,7,13,19}
def binFindUp(arr, key): low = 0 high = len(arr) -1 while(low < high): print "%d,%d" % (low, high) mid = (low + high) / 2 if (arr[mid] <= key): low = mid else: high = mid - 1 return low def binFindDown(arr, key): low = 0 high = len(arr) -1 while(low < high): print "%d,%d" % (low, high) mid = (low + high) / 2 if (arr[mid] >= key): high = mid else: low = mid + 1 return high 2. 如何计算两个有序整形数组的交集,比如： a=0,1,2,3,4 b=1,3,5,7,9
由于求交集，交集一定是集合之间的运算，所以说a,b是集合，即它们本身不含重复元素，参考归并排序的方法即可。
def interset(a, b) i = 0 j = 0 c = [] while i < len(a) and b < len(b): if a[i] == b[j]: c.add(a[i]) i++ j++ else if a[i] > b[j]: i++ else: j++ return c3. 求最长公共子串
def lcs(str1, str2): dp = [[0 for col in range(len(str2) + 1)] for row in range(len(str1) + 1)] for i in range(1, len(str1) + 1): for j in range(1, len(str2) + 1): dp[i][j] = max(dp[i-1][j], dp[i][j-1], dp[i-1][j-1] + (1 if str1[i-1] == str2[j-1] else 0)) return dp[len(str1)][len(str2)]4.字符串编辑距离
def leven(str1, str2): dp = [[0 for col in range(len(str2) + 1)] for row in range(len(str1) + 1)] for i in range(0, len(str1) + 1): for j in range(0, len(str2) + 1): if i == 0 or j == 0: dp[i][j] = i + j else: dp[i][j] = min(dp[i-1][j] + 1, dp[i][j-1] + 1, dp[i-1][j-1] + (0 if str1[i-1] == str2[j-1] else 1)) return dp[len(str1)][len(str2)] 5.求无序数组的第n小值。如果排序再求，那么是nlgn的时间复杂度，下面是一个n的线性复杂度，可用来求最大值，最小值，中位数等。
def npos(lst, n): index = random.randint(0, len(lst) - 1) big = [l for l in lst if l > lst[index]] little = [l for l in lst if l < lst[index]] equal = [l for l in lst if l == lst[index]] if len(little) > n - 1: return npos(little, n) if len(little) <= n - 1: if len(little) + len(equal) >= n: return lst[index] return npos(big, n - len(little) - len(equal))
