【LeetCode】303. Range Sum Query - Immutable 解题报告（Python）

id： fuxuemingzhu

个人博客： ​​​http://fuxuemingzhu.cn/​​​

目录

• • 
    
• ​​题目描述​​
• ​​解题方法​​

• ​​保存累积和​​

• ​​日期​​
  • 
    

题目地址：​​https://leetcode.com/problems/range-sum-query-immutable/description/​​

题目描述

Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.

Example:
Given nums = [-2, 0, 3, -5, 2, -1]

sumRange(0, 2) -> 1
sumRange(2, 5) -> -1
sumRange(0, 5) -> -3

Note:

• • 
    
• You may assume that the array does not change.
• There are many calls to sumRange function.
  • 
    

解题方法

保存累积和

可以直接用切片求和的方法做，也能A，但是效率太慢。

下面这个方式可以先把sums求出来，然后再调用的时候直接右边的sums-左边的sums即可得到结果。

class NumArray(object):

    def __init__(self, nums):
        """
        :type nums: List[int]
        """
        self.sums = [0] * len(nums)
        total = 0
        for i, num in enumerate(nums):
            total += num
            self.sums[i] = total


    def sumRange(self, i, j):
        """
        :type i: int
        :type j: int
        :rtype: int
        """
        if i == 0:
            return self.sums[j]
        else:
            return self.sums[j] - self.sums[i - 1]


# Your NumArray object will be instantiated and called as such:
# obj = NumArray(nums)
# param_1 = obj.sumRange(i,j)

如果多用一个元素放在开头，那么上面的这个代码可以简化。

class NumArray(object):

    def __init__(self, nums):
        """
        :type nums: List[int]
        """
        N = len(nums)
        self.sums = [0] * (N + 1)
        for i in range(1, N + 1):
            self.sums[i] = self.sums[i - 1] + nums[i - 1]

    def sumRange(self, i, j):
        """
        :type i: int
        :type j: int
        :rtype: int
        """
        return self.sums[j + 1] - self.sums[i]


# Your NumArray object will be instantiated and called as such:
# obj = NumArray(nums)
# param_1 = obj.sumRange(i,j)

日期

2018 年 2 月 4 日

2018 年 11 月 24 日 —— 周六快乐