[leetcode] 303. Range Sum Query - Immutable

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.
解法一：

直观上如果每次去循环相加肯定不行。这里思路是根据前i个元素的和建立直方图，如果求i-j元素之和就是dp[j]-dp[i-1]. 注意到i==0是个corner case，这时候返回值应为dp[j]。

class NumArray {
public:
NumArray(vector<int> &nums) {
dp = nums;
for(int i=1; i<nums.size();++i){
dp[i] += dp[i-1];
}

}

int sumRange(int i, int j) {
return i==0?dp[j]:dp[j]-dp[i-1];
}
private:
vector<int> dp;
};

// Your NumArray object will be instantiated and called as such:
// NumArray numArray(nums);
// numArray.sumRange(0, 1);
// numArray.sumRange(1, 2);