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:

You may assume that the array does not change.

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);