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.

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解法一：

O(1) init, O(n) query

public class NumArray {

private int[] nums;
private int sum;
public NumArray(int[] nums) {
this.nums = nums;
this.sum = 0;
}

public int sumRange(int i, int j) {
sum = 0;      //保证immutable
for(int k = i;k<=j;k++)
{
sum = sum + nums[k];
}

return sum;
}
}

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

解法二：

O(n) init, O(1) query

public class NumArray {

int[] nums;

public NumArray(int[] nums) {
for(int i = 1; i < nums.length; i++)
nums[i] += nums[i - 1];

this.nums = nums;
}

public int sumRange(int i, int j) {
if(i == 0)
return nums[j];

return nums[j] - nums[i - 1];
}
}

reference: https://discuss.leetcode.com/topic/29194/java-simple-o-n-init-and-o-1-query-solution