Range Sum Query

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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 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