943. Range Sum Query - Immutable

题解

　　一开始我用dp[i][j]来存i到j的和，然后直接返回dp[i][j]即可，时间复杂度n*n，但提交爆内存了，那就是说可以一维数组解决咯，于是想到用dp[i]代表前i个数字的和，找i j之间就是dp[j] - dp[i-1]这里要注意是i-1，否则会多减掉一个数字，这样就出答案

class NumArray(object):

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

    def sumRange(self, i, j):
        """
        :type i: int
        :type j: int
        :rtype: int
        """
        #print self.book,self.book[j] , self.book[i]
        
        return self.book[j] - self.book[i-1]