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

Algorithm

每日一刷-Range-Sum-Query-2D-Immutable

[每日一刷] (Range Sum Query 2D Immutable)


Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2).

Example:
Given matrix = [
[3, 0, 1, 4, 2],
[5, 6, 3, 2, 1],
[1, 2, 0, 1, 5],
[4, 1, 0, 1, 7],
[1, 0, 3, 0, 5]
]

sumRegion(2, 1, 4, 3) -> 8
sumRegion(1, 1, 2, 2) -> 11
sumRegion(1, 2, 2, 4) -> 12

解题思路


这题思路和一维的时候一致，可以使用动态规划也可以使用线段树。就这题来说使用动态规划会简单很多。
思路就是用dp[i][j]表示从(0, 0)到(i, j)的和

Python:

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#!/usr/bin/env 
# -*- coding:utf-8 -*-
"""
时间复杂度 Query O(1), Build O(n^2)
"""
class NumMatrix(object):
    def __init__(self, matrix):
        """
        initialize your data structure here.
        :type matrix: List[List[int]]
        """
        if matrix == []:
            return
        r = len(matrix)
        c = len(matrix[0])
        self.dp = [[0 for i in xrange(c+1)] for j in xrange(r+1)]
        for i in xrange(1, r+1):
            for j in xrange(1, c+1):
                self.dp[i][j] = self.dp[i-1][j] + self.dp[i][j-1] - self.dp[i-1][j-1] + matrix[i-1][j-1]
    def sumRegion(self, row1, col1, row2, col2):
        """
        sum of elements matrix[(row1,col1)..(row2,col2)], inclusive.
        :type row1: int
        :type col1: int
        :type row2: int
        :type col2: int
        :rtype: int
        """
        return self.dp[row2+1][col2+1] - self.dp[row1][col2+1] - self.dp[row2+1][col1] + self.dp[row1][col1]