[LeetCode] 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.

解题思路

考虑到要多次调用sumRange()函数，因此需要把结果先存起来，调用时就可以直接返回了。最开始考虑的是用dp[i][j]来直接存储i到j之间元素的和，但是内存超出限制。于是考虑用dp[i]来存储0到i之间元素的和，0到j的和减去0到i-1的和即为所求。

实现代码

// Runtime: 3 ms
public class NumArray {
    private int[] dp;
    public NumArray(int[] nums) {
        dp = new int[nums.length];
        int sum = 0;
        for (int i = 0; i < nums.length; i++) {
            sum += nums[i];
            dp[i] = sum;
        }
    }

    public 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 = new NumArray(nums);
// numArray.sumRange(0, 1);
// numArray.sumRange(1, 2);