【一天一道LeetCode】#303.Range Sum Query - Immutable
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(一)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.
(二)解题
题目大意:给定一个数组,计算它的区间和
解题思路:本题给出了NumArray的构造函数,区间和应该再构造函数内就已经计算好。
class NumArray {
public:
NumArray(vector<int> &nums) {
int size = nums.size();
int sum = 0;
for(int i = 0 ; i < size ; i++){
vec.push_back(sum);//vec里面存放第0~i-1位数的和。
sum+=nums[i];
}
vec.push_back(sum);//第0~i位的和需要压入vector
}
int sumRange(int i, int j) {
return vec[j+1] - vec[i];//直接计算区间和
}
public:
vector<int> vec;
};
// Your NumArray object will be instantiated and called as such:
// NumArray numArray(nums);
// numArray.sumRange(0, 1);
// numArray.sumRange(1, 2);
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