Algorithm

两个指针-TwoPointer

[两个指针] (Two Pointer)

两个指针的题目主要分成三种情况：

a.对撞型

  1. 2Sum 类
    例题: Trapping Water
    一个指针i指向数组头，一个指针j指向数组尾。

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    while i != j:
      if num[i] < num[j]:
        i += 1
      elif num[i] > num[j]:
        j -= 1
      else:
        i += 1

  2. Partition类
    例题: 将一个数组分成两部分，一部分小与k，一部分大于k
    一个指针i指向数组头，一个指针j指向数组尾。i表示在i之前的元素都小于k，j表示在j之后的元素都大于k。

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    while i != j:
      while num[i] < k:
        i += 1
      while num[j] > k:
        j -= 1
      if i > j:
        break
      else:
        swap(i, j)

总结
通过对撞型指针优化算法，根本上其实要证明不用扫描多余状态

b.前向型

  1. 窗口类
    例题: Minimum Window Substring
    窗口类指针模板j

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    j = 0
    for i in xrange(n):
      while j < n: 
        if 满足条件:
          j += 1
          更新状态
        else 不满足条件:
          break

  2. 快慢类
    两个指针i, j的pace不同，可以用来探测一个链表是否有回路

总结
窗口类的问题根本上其实也是要证明不用扫描多余状态
快慢类的问题感觉和经常见的烧蜡烛的那种智力题很像。或者是环绕一圈来找list有没有cycle。

c.两个数组，两个指针

  1. 并行
    例题: Merge Two Sorted Array
    单纯的是否两个指针
Algorithm

每日一刷-Count of Smaller Numbers After Self

[每日一刷] (Count of Smaller Numbers After Self)


You are given an integer array nums and you have to return a new counts array.
The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i].

Example:
Given nums = [5, 2, 6, 1]
Return the array [2, 1, 1, 0]

解题思路


这题其实又是一个有关区间的题，每次都找当前值后面有多少个比它小的数。因此马上可以想到用线段树。
这道题其实就是值线段树的简单应用，线段树结点中存储一个区间内有多少个值，然后每次查询后修改线段树。
这题主要就是各种边界条件比较多，下次要注意。

Python:

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"""
时间复杂度 O(nlog(n))
"""
class SegmentTree(object):
    def __init__(self, d, nums, left, right):
        self.left, self.right = left, right
        self.val = 0
        self.mid = self.left + (self.right - self.left) / 2
        self.c1 = self.c2 = None
        if self.left == self.right:
            self.val = d[self.left]
        else:
            self.c1 = SegmentTree(d, nums, self.left, self.mid)
            self.c2 = SegmentTree(d, nums, self.mid+1, self.right)
            self.val = self.c1.val + self.c2.val
            
    def query(self, right):
        if right < self.left:
            return 0
        if self.right == self.left:
            return self.val
        if self.right == right:
            return self.val
        else:
            if right <= self.mid:
                return self.c1.query(right)
            else:
                return self.c1.query(self.mid) + self.c2.query(right)
    
    def modify(self, index):
        if self.left == self.right and self.left == index:
            self.val -= 1
        else:
            if index <= self.mid:
                self.c1.modify(index)
            else:
                self.c2.modify(index)
            self.val = self.c1.val + self.c2.val
        
class Solution(object):
    def countSmaller(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        if nums == []:
            return []
        left, right = min(nums), max(nums)
        for i in xrange(len(nums)): # deal with numbers below zero.
            nums[i] -= left
        right -= left
        d = [0 for i in xrange(right+1)]
        for i in nums:
            d[i] += 1
        tree = SegmentTree(d, nums, 0, right) # build segment tree.
        res = []
        for i in nums:
            temp = tree.query(i-1)
            res.append(temp)
            tree.modify(i) # modify the tree after each query
        return res