LeetCode 303. Range Sum Query - Immutable

原题链接在这里：https://leetcode.com/problems/range-sum-query-immutable/

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:

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

题解： 

因为这个函数会被多次调用，每次都算一遍会花很多时间, 所以用dp数组来保存过去的值.

dp[i] 表示 nums[0]到nums[i-1]的和，如此做是为了求sumRange时不需要额外讨论 i == 0的情况。

Time Complexity: NumArray, O(n). sumRange, O(1). n = nums.length.

Space: O(n).

AC Java:

public class NumArray {
    private int [] dp;
    public NumArray(int[] nums) {
        dp = new int[nums.length+1]; //dp[i]表示从nums[0]到nums[i-1]的和
        for(int i = 1; i<=nums.length; i++){
            dp[i] = dp[i-1] + nums[i-1];
        }
    }

    public int sumRange(int i, int j) {
        return dp[j+1] - dp[i];
    }
}

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

类似Range Sum Query - Mutable, Range Sum Query 2D - Immutable.