303. Range Sum Query - Immutable

1.Question

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.

2.Code

class NumArray {
private:
vector<int> dp;
public:
NumArray(vector<int> &nums) {
dp = nums;
for(int i = 1, size = nums.size(); i < size; i++)
dp[i] += dp[i-1]; //实际上也是dp[i] = dp[i-1] + nums[i]
}

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

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

3.Note

a.
这个题里我们要注意，sumRange() 会多次调用，如果直接遍历求和会超时。那么，我们用动态规划，用数组dp 记录0~i的和，表示为dp[i]。这样如果i == 0则sumRange(i, j)返回dp[j]即可， i != 0时， i~j 的和可以用dp[j] - dp[i-1]表示。

b.
dp是个vector，初始化分配空间很重要。否则会出现RE.