leetcode解题之303# Range Sum Query - Immutable Java版 (多次计算数组内任意两个下标之间的和)

303. Range Sum Query - Immutable

Given an integer array nums, find the sum of the elements between indicesi 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.

给定一个整数序列，求指定子序列和。若每次遍历求解，是不通过的。

我们可以存储子序列和，每个下标处的值为[0,i]的所有元素和;

那么[i,j]子序列和=sum[j]−sum[i−1]；

/**
 * Your NumArray object will be instantiated and called as such:
 * NumArray obj = new NumArray(nums);
 * int param_1 = obj.sumRange(i,j);
 */

public class NumArray {
	private int[] sum;

	public NumArray(int[] nums) {
		if (nums.length > 0) {
			sum = new int[nums.length];
			sum[0] = nums[0];
			for (int i = 1; i < nums.length; i++) {
				sum[i] = sum[i - 1] + nums[i];
			}
		}
	}

	public int sumRange(int i, int j) {
		// 注意处理i=0的情况
		if (i == 0)
			return sum[j];
		else
			return sum[j] - sum[i - 1];
	}
}