题目:
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.
解释:
当然可以直接用求和函数,但是特别慢。其实这种题目一般需要用一个数组储存前n项和,然后在求rangeSum的时候直接相减即可,前n项和是一个很经典的方法,如sums[i]
保存包括nums[i]
的前i+1
项和。
直接用求和函数,python代码:
class NumArray:
def __init__(self, nums):
"""
:type nums: List[int]
"""
self.nums=nums
def sumRange(self, i, j):
"""
:type i: int
:type j: int
:rtype: int
"""
return sum(self.nums[i:j+1])
# Your NumArray object will be instantiated and called as such:
# obj = NumArray(nums)
# param_1 = obj.sumRange(i,j)
前n项和解法,python代码:
class NumArray(object):
def __init__(self, nums):
"""
:type nums: List[int]
"""
n=len(nums)
self.stepsum=[0]*n
if n>0:
self.stepsum[0]=nums[0]
for i in xrange(1,n):
self.stepsum[i]=self.stepsum[i-1]+nums[i]
def sumRange(self, i, j):
"""
:type i: int
:type j: int
:rtype: int
"""
return self.stepsum[j]-(self.stepsum[i-1] if i >0 else 0)
# Your NumArray object will be instantiated and called as such:
# obj = NumArray(nums)
# param_1 = obj.sumRange(i,j)
c++代码:
class NumArray {
public:
vector<int>sums;
NumArray(vector<int> nums) {
int n=nums.size();
sums.assign(n,-1);
if (n>0)
sums[0]=nums[0];
for (int i=1;i<nums.size();i++)
{
sums[i]=sums[i-1]+nums[i];
}
}
int sumRange(int i, int j) {
return sums[j]- (i>0?sums[i-1]:0);
}
};
/**
* Your NumArray object will be instantiated and called as such:
* NumArray obj = new NumArray(nums);
* int param_1 = obj.sumRange(i,j);
*/
总结:
c++中给一个已经声明好的vector 赋值:
vector<int> sums;
sums.assign(nums.begin(),nums.end());
(这个比下面那句速度更快)
sums=vector<int>(nums.begin(),nums.end());