[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:

You may assume that the array does not change.

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);