LeetCode No.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:

1. You may assume that the array does not change.
2. There are many calls to sumRange function.

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題目鏈接：https://leetcode.com/problems/range-sum-query-immutable/

題目大意：給定一個數組，多次使用sumRange(i,j)返回數組中從i到j的元素之和。

思路：先用一個和nums一樣大的數組sums存儲nums的和，其中sums[i] = sums[i-1] + nums[i]，時間複雜度爲O(N)，空間複雜度爲O(N)。最後計算sumRange(i,j)時，結果就是sums[j] - sums[i] + nums[i]sums[j] - sums[i] + nums[i]，這樣的時間複雜度爲O(1)，因爲多次使用sumRange()函數，所以這樣做能大大減少算法運算時間。

參考代碼：

class NumArray {
public:
    NumArray(vector<int> &nums) {
        this -> nums = nums ;
        int n = nums.size() ;
        if ( n )
        {
            sums = vector <long long> ( n ) ; 
            sums[0] = nums[0] ;
            for ( int i = 1 ; i < n ; i ++ )
                sums[i] = sums[i-1] + nums[i] ;
        }
    }

    int sumRange(int i, int j) {
        return sums[j] - sums[i] + nums[i] ;
    }
private:
    vector <long long> sums ;
    vector <int> nums ;
};


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