LeetCode -- Range Sum Query

geQuestion:

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.

 

Analysis:

给出一个整数数组，得出i与j之间的数字之和（i<=j）。你可以假设数组不会改变，并且他们可以调用多次sunRange函数。

 

Answer：

解法一：

public class NumArray {
    int[] num;
    public NumArray(int[] nums) {
        this.num = nums;
    }
    public int sumRange(int i, int j) {
        int res = 0;
        for(int k=i; k<j+1; k++) {
            res += num[k];
        }
        return res;
    }
    
}


// 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：由于题目中说会多次调用该函数，计算某个区间内的和，如果按照上面的解法，当数组过大时，会导致多次重复计算和，因此我们可以想办法存储中间计算的结果来简化计算。 方法就是一次存储从第0个元素到第i个元素计算的中间结果，当求某个区间的和时，直接用大区间减去小区间的和即可。

public class NumArray {

    int[] num;
    public NumArray(int[] nums) {
        if(nums == null)
            num = null;
        else if(nums.length == 0)
            num = nums;
        else {
            num = new int[nums.length];
            num[0] = nums[0];
            for(int i=1; i<nums.length; i++) {
                num[i] = nums[i] + num[i-1];
            }
            for(int i=0; i<num.length; i++) 
                System.out.print(num[i] + " * ");
            System.out.println();
        }
        

    }
    public int sumRange(int i, int j) {
        if(i == 0)
            return num[j];
        else return num[j] - num[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);