第一周作业#03

Week_01/id_3/NOTE.md

@@ -1 +1,14 @@
 # 学习笔记
+
+## 小技巧
+python的数组下标访问比hash快多了
+
+l1, l2 = l2, l1
+这个语法不但简单，而且思路其实适用于任何语言。
+两个变量没有顺序要求，我们可以通过预处理的方式，强制变成有先后关系，这就避免了维护一个中间标量的尴尬
+
+python的异或实现 bool(n1) != bool(n2) js也可以参考
+
+
+DP 子问题有最优解 -> 全局最优解 找到最优子结构即可向上递推
+递归 可以遍历所有解，这个是DP做不到的

Week_01/id_3/array/15/LeetCode_15_3_v1.py

@@ -0,0 +1,54 @@
+def three_sum(nums):
+    r = []
+    if not nums or len(nums) < 3:
+        return r
+
+    cache = {}
+    dset = set()
+
+    new_nums = []
+
+    for n in nums:
+        count = 1 + cache.get(n, 0)
+        cache[n] = count
+        if n and count <= 2:
+            new_nums.append(n)
+
+    zero_count = cache.get(0, 0)
+    if zero_count >= 3:
+        r.append([0, 0, 0])
+
+    if zero_count > 0:
+        new_nums.append(0)
+
+    nums = new_nums
+    nums.sort()
+
+    length = len(nums)
+    for i in range(length):
+        n1 = nums[i]
+        # n1 >= 0 时说明已经遍历把有效的数据都进行遍历，继续遍历已经发生重复
+        if n1 >= 0:
+            break
+        for j in range(i+1, length):
+            n2 = nums[j]
+            if n1 == 0 and n2 == 0 and cache[0] < 3:
+                continue
+            s = -(n1 + n2)
+
+            # 迭代顺序是从小到大的
+            # 当-sum(n1,n2)的结果小于n2时 说明已经迭代出现了重复 -sum必然在之前已经迭代过
+            if s < n2:
+                break
+
+            if (s == n1 or s == n2) and cache.get(s, 0) == 1:
+                continue
+
+            if cache.get(s, 0):
+                arr = [n1, n2, s]
+                ds = '%s,%s' % (n1, n2)
+                if ds not in dset:
+                    dset.add(ds)
+                    r.append(arr)
+
+    return r

Week_01/id_3/array/15/LeetCode_15_3_v2.py

@@ -0,0 +1,6 @@
+def three_sum(nums):
+    nums, r = sorted(nums), set()
+    for i in [i for i in range(len(nums)-2) if i < 1 or nums[i] > nums[i-1]]:
+        d = {-nums[i]-n: j for j, n in enumerate(nums[i + 1:])}
+        r.update([(nums[i], n, -nums[i]-n) for j, n in enumerate(nums[i+1:]) if n in d and d[n] > j])
+    return list(map(list, r))

Week_01/id_3/array/15/test_15.py

@@ -0,0 +1,24 @@
+import lc_15_v1
+import lc_15_v2
+
+f = lc_15_v1.three_sum
+# f = lc_15_v2.three_sum
+
+
+def check(nums, r):
+    _r = f(nums)
+    print(_r, r, _r == r)
+
+
+check([-1, 0, 1, 2, -1, -4],
+      [
+          [-1, 0, 1],
+          [-1, -1, 2]
+      ])
+
+
+check([0, 0], [])
+check([0, 0, 0], [[0, 0, 0]])
+
+check([-4, -2, -2, -2, 0, 1, 2, 2, 2, 3, 3, 4, 4, 6, 6],
+      [[-4, -2, 6], [-4, 0, 4], [-4, 1, 3], [-4, 2, 2], [-2, -2, 4], [-2, 0, 2]])

Week_01/id_3/array/189/LeetCode_189_3_v1.py

@@ -0,0 +1,27 @@
+# 最大公约数
+# 每次数组挪动一步，挪动k次，毫无疑问是可以实现的，但是时间复杂度 O(kn) 是无法接受的
+# 如果每次移动的是nums[i]和nums[i+k]，很可能无法遍历所有数据，例如len(nums)==9 k==3，此时只会在0,3,6三个下标之间进行移动
+# 根据规律总结 如len(nums)与k的最小公倍数 是每次进行k偏移交换的循环
+# 也就是len(nums)与k的最大公约数gcd会是产生循环迭代的次数。例如len(nums)==9 k==3 gcd==3 下标每次偏移3，3次就会出现循环。
+# 所以需要 len(nums)/gcd 次就可以交换所有的数据
+
+import math
+
+
+def rotate(nums, k):
+    if not nums:
+        return
+
+    length = len(nums)
+    k = k % length
+
+    if not k:
+        return
+
+    gcd = math.gcd(length, k)
+
+    for offset in range(gcd):
+        tmp = nums[offset]
+        for i in range(1, int(length/gcd) + 1):
+            index = (i*k + offset) % length
+            nums[index], tmp = tmp, nums[index]

Week_01/id_3/array/189/LeetCode_189_3_v2.py

@@ -0,0 +1,24 @@
+# 交换法
+# 1,2,3,4,5,6,7 进行3位置的移动 等价于 list[0:4]+list[0:3] 两段数据的位置交换
+# 位置交换的原地极简方案就是两次反转 [1,2,3,4,5,6,7] 反转0:4-> [4,3,2,1,5,6,7] 反转4:7-> [4,3,2,1,7,6,5] 全部反转-> [5,6,7,1,2,3,4]
+
+
+def rotate(nums, k):
+    if not nums:
+        return
+
+    length = len(nums)
+    if length == 1:
+        return
+
+    k = k % length
+
+    tail = length - k - 1
+    for i in range(int((length-k)/2)):
+        nums[i], nums[tail-i] = nums[tail-i], nums[i]
+
+    tail = -k
+    for i in range(int(k/2)):
+        nums[-i-1], nums[tail+i] = nums[tail+i], nums[-i-1]
+
+    nums.reverse()

Week_01/id_3/array/189/test_189.py

@@ -0,0 +1,24 @@
+import lc_189_v1
+import lc_189_v2
+
+# f = lc_189_v1.rotate
+f = lc_189_v2.rotate
+
+
+def check(params, k, arr):
+    _p = list(params)
+    f(params, k)
+    print(_p, params == arr)
+
+
+check([1, 2, 3, 4, 5, 6, 7], 3, [5, 6, 7, 1, 2, 3, 4])
+check([1, 2, 3, 4, 5, 6], 3, [4, 5, 6, 1, 2, 3])
+check([1, 2, 3], 7, [3, 1, 2])
+check([-1], 2, [-1])
+check(
+    [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31,
+     32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54],
+    45,
+    [10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38,
+     39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 1, 2, 3, 4, 5, 6, 7, 8, 9]
+)

Week_01/id_3/array/26/LeetCode_26_3_v1.py

@@ -0,0 +1,12 @@
+def remove_duplicates(nums):
+    if not nums:
+        return 0
+
+    i = 1
+    cur = nums[0]
+    for n in nums:
+        if n != cur:
+            nums[i] = cur = n
+            i += 1
+
+    return i

Week_01/id_3/array/26/test_26.py

@@ -0,0 +1,12 @@
+import lc_26_v1
+
+f = lc_26_v1.remove_duplicates
+
+
+def check(params, length, arr):
+    _p = list(params)
+    print(_p, f(params) == length and params[:length] == arr)
+
+
+check([], 0, [])
+check([1, 1, 2, 3], 3, [1, 2, 3])

Week_01/id_3/array/88/LeetCode_88_3_v1.py

@@ -0,0 +1,16 @@
+# nums1 m -> m+n 位置是剩余空间 如果按从小到大排序会出现复杂的冲突
+# 而从大到小排列， 最大的元素移动到剩余空间的末端，则可以避免出现冲突
+
+
+def merge(nums1, m, nums2, n):
+    tail = m + n
+    for i in range(tail):
+        if n == 0:
+            break
+        tail -= 1
+        if m > 0 and nums1[m-1] > nums2[n-1]:
+            m -= 1
+            nums1[tail] = nums1[m]
+        else:
+            n -= 1
+            nums1[tail] = nums2[n]

Week_01/id_3/array/88/LeetCode_88_3_v2.py

@@ -0,0 +1,5 @@
+# 题目并未要求原地排序 所以弄个简单的
+
+
+def merge(nums1, m, nums2, n):
+    nums1[:m+n] = sorted(nums1[:m] + nums2[:n])

Week_01/id_3/array/88/test_88.py

@@ -0,0 +1,13 @@
+import lc_88_v1
+import lc_88_v2
+
+f = lc_88_v1.merge
+# f = lc_88_v2.merge
+
+
+def check(nums1, m, nums2, n, r):
+    f(nums1, m, nums2, n)
+    print(nums1, nums1 == r)
+
+
+check([1, 2, 3, 0, 0, 0], 3, [2, 5, 6], 3, [1, 2, 2, 3, 5, 6])

Week_01/id_3/bs/441/LeetCode_441_3_v1.py

@@ -0,0 +1,10 @@
+def arrange_coins(n):
+    if n <= 0:
+        return 0
+    s = 0
+    k = 0
+    while s <= n:
+        k += 1
+        s += k
+
+    return k - 1

Week_01/id_3/bs/441/LeetCode_441_3_v2.py

@@ -0,0 +1,16 @@
+# 此题超时了 有待用二分优化
+
+
+def arrange_coins(n):
+    if n <= 0:
+        return 0
+    k = 1
+
+    while True:
+        s = int(((1 + k) * k)/2)
+        print(s, k)
+        if s == n:
+            return k
+        if s > n:
+            return k - 1
+        k += 1

Week_01/id_3/bs/441/test_441.py

@@ -0,0 +1,15 @@
+import lc_441_v1
+import lc_441_v2
+
+# f = lc_441_v1.arrange_coins
+f = lc_441_v2.arrange_coins
+
+
+def check(n, r):
+    _r = f(n)
+    print(_r, r, _r == r)
+
+
+check(5, 2)
+check(8, 3)
+check(2146533481, 65521)

Week_01/id_3/bs/bs_basis.py

@@ -0,0 +1,90 @@
+def bs(nums, x):
+    if not nums:
+        return -1
+    length = len(nums)
+    low = 0
+    high = length - 1
+    while low <= high:
+        mid = low + int((high - low)/2)
+        v = nums[mid]
+        if v == x:
+            return mid
+        if v > x:
+            high = mid - 1
+        else:
+            low = mid + 1
+
+    return -1
+
+
+def bs_first(nums, x):
+    if not nums:
+        return -1
+    length = len(nums)
+    low = 0
+    high = length - 1
+    while low <= high:
+        mid = low + int((high - low)/2)
+        v = nums[mid]
+        if v >= x:
+            high = mid - 1
+        else:
+            low = mid + 1
+
+    if low < length and nums[low] == x:
+        return low
+
+    return -1
+
+
+def bs_last(nums, x):
+    low = 0
+    high = len(nums) - 1
+    while low <= high:
+        mid = low + (high - low)//2
+        v = nums[mid]
+        if v > x:
+            high = mid - 1
+        else:
+            low = mid + 1
+    if high >= 0 and nums[high] == x:
+        return high
+    return -1
+
+
+def bs_gte(nums, x):
+    low = 0
+    high = len(nums) - 1
+    while low <= high:
+        mid = low + (high - low)//2
+        v = nums[mid]
+        if v >= x:
+            high = mid - 1
+        else:
+            low = mid + 1
+
+    return low
+
+
+# print(bs([8, 11, 19, 23, 27, 33, 45, 55, 67, 98], 19) == 2)
+# print(bs([8, 11, 19, 23, 27, 33, 45, 55, 67, 98], 33) == 5)
+# print(bs([8, 11, 19, 23, 27, 33, 45, 55, 67, 98], 18) == -1)
+# print(bs([8, 8, 11, 11, 19, 23, 27, 33, 45, 55, 67, 98], 19) == 4)
+#
+# print(bs_first([8, 8, 8, 11, 11, 19, 23, 27, 33, 45, 55, 67, 98], 11) == 3)
+# print(bs_first([8, 8, 8, 11, 11, 19, 23, 27, 33, 45, 55, 67, 98], 8) == 0)
+# print(bs_first([8, 8, 11, 11, 19, 23, 27, 33, 45, 55, 67, 98], 18) == -1)
+# print(bs_first([8, 8, 11, 11, 19, 23, 27, 33, 45, 55, 67, 98], 188) == -1)
+# print(bs_first([8, 8, 11, 11, 19, 23, 27, 33, 45, 55, 67, 98], 1) == -1)
+
+print(bs_last([8, 8, 8, 11, 11, 19, 23, 27, 33, 45, 55, 67, 98], 11) == 4)
+print(bs_last([8, 8, 8, 11, 11, 19, 23, 27, 33, 45, 55, 67, 98], 8) == 2)
+print(bs_last([8, 8, 11, 11, 19, 23, 27, 33, 45, 55, 67, 98], 18) == -1)
+print(bs_last([8, 8, 11, 11, 19, 23, 27, 33, 45, 55, 67, 98], 188) == -1)
+print(bs_last([8, 8, 11, 11, 19, 23, 27, 33, 45, 55, 67, 98], 1) == -1)
+
+# print(bs_gte([8, 8, 11, 11, 19, 23, 27, 33, 45, 55, 67, 98], 17) == 4)
+# print(bs_gte([8, 8, 11, 11, 19, 23, 27, 33, 45, 55, 67, 98], 19) == 4)
+# print(bs_gte([8, 8, 11, 11, 19, 23, 27, 33, 45, 55, 67, 98], 100) == 12)
+# print(bs_gte([8, 8, 11, 11, 19, 23, 27, 33, 45, 55, 67, 98], 8) == 0)
+# print(bs_gte([8, 8, 11, 11, 19, 23, 27, 33, 45, 55, 67, 98], 7) == 0)

Week_01/id_3/hash/242/LeetCode_242_3_v1.py

@@ -0,0 +1,16 @@
+def is_anagram(s, t):
+    if len(s) != len(t):
+        return False
+
+    arr = [0] * 26
+    m = ord('a')
+    for c in s:
+        arr[ord(c) % m] += 1
+
+    for c in t:
+        i = ord(c) % m
+        if arr[i] == 0:
+            return False
+        arr[i] -= 1
+
+    return True