303.区域和检索-数组不可变.py

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# @lc app=leetcode.cn id=303 lang=python
#
# [303] 区域和检索 - 数组不可变
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# https://leetcode-cn.com/problems/range-sum-query-immutable/description/
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# algorithms
# Easy (56.63%)
# Likes:    105
# Dislikes: 0
# Total Accepted:    19.6K
# Total Submissions: 34.2K
# Testcase Example:  '["NumArray","sumRange","sumRange","sumRange"]\n[[[-2,0,3,-5,2,-1]],[0,2],[2,5],[0,5]]'
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# 给定一个整数数组  nums，求出数组从索引 i 到 j  (i ≤ j) 范围内元素的总和，包含 i,  j 两点。
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# 示例：
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# 给定 nums = [-2, 0, 3, -5, 2, -1]，求和函数为 sumRange()
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# sumRange(0, 2) -> 1
# sumRange(2, 5) -> -1
# sumRange(0, 5) -> -3
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# 说明:
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# 
# 你可以假设数组不可变。
# 会多次调用 sumRange 方法。
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# 
#

# @lc code=start
class NumArray(object):

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


    def sumRange(self, i, j):
        """
        :type i: int
        :type j: int
        :rtype: int
        """
        if i==0:return self.sums[j]
        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)
# @lc code=end