题目描述
给定一个整数数组 nums
,处理以下类型的多个查询:
- 计算索引
left
和right
(包含left
和right
)之间的nums
元素的 和 ,其中left <= right
实现 NumArray
类:
NumArray(int[] nums)
使用数组nums
初始化对象int sumRange(int i, int j)
返回数组nums
中索引left
和right
之间的元素的 总和 ,包含left
和right
两点(也就是nums[left] + nums[left + 1] + ... + nums[right]
)
示例 1:
输入:
["NumArray", "sumRange", "sumRange", "sumRange"]
[[[-2, 0, 3, -5, 2, -1]], [0, 2], [2, 5], [0, 5]]
输出:
[null, 1, -1, -3]
解释:
NumArray numArray = new NumArray([-2, 0, 3, -5, 2, -1]);
numArray.sumRange(0, 2); // return 1 ((-2) + 0 + 3)
numArray.sumRange(2, 5); // return -1 (3 + (-5) + 2 + (-1))
numArray.sumRange(0, 5); // return -3 ((-2) + 0 + 3 + (-5) + 2 + (-1))
提示:
1 <= nums.length <= 104
-105 <= nums[i] <= 105
0 <= i <= j < nums.length
- 最多调用
104
次sumRange
方法
解决方案:
1、利用状态方程求前缀和,一次遍历即可:sums[i + 1] = sums[i] + nums[i];
2、C++库函数:partial_sum(nums.begin(),nums.end(),sums.begin()+1);
函数源码:
方案一:
cppclass NumArray { public: vector<int> sums; NumArray(vector<int>& nums) { int n = nums.size(); sums.resize(n + 1); for (int i = 0; i < n; i++) { sums[i + 1] = sums[i] + nums[i]; } } int sumRange(int i, int j) { return sums[j + 1] - sums[i]; } };
方案二:
cppclass NumArray { vector<int>sums; public: NumArray(vector<int>& nums) : sums(nums.size()+1,0){ partial_sum(nums.begin(),nums.end(),sums.begin()+1); } int sumRange(int left, int right) { return sums[right+1]-sums[left]; } }; /** * Your NumArray object will be instantiated and called as such: * NumArray* obj = new NumArray(nums); * int param_1 = obj->sumRange(left,right); */