Range Sum Query - Immutable

题目描述：

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.

这个题说 sumRange会被调用很多次 ，所以如果用下面的方法就会超时：

public int[] nums;

public Solution303(int[] nums) {
    this.nums = nums;
}

public int sumRange(int i, int j) {
    int result = 0;
    for (int counter = i; counter <= j; counter++) {
        result += nums[counter];
    }
    return result;
}

所以我们要做的是在sumRange中尽可能的时间复杂度是O(1)。sum(i,j)=sum(0,j)-sum(0,i)

正确方法如下：

public class NumArray {

    List<Integer> list=new ArrayList<Integer>();
	public NumArray(int[] nums) {
        for(int i=0;i<nums.length;i++){
        	list.add(nums[i]);
        }
    }

    public int sumRange(int i, int j) {
        int sum=0;
    	for(int k=i;k<=j;k++){
        	sum+=list.get(k);
        }
        return sum;
    }
}