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:

You may assume that the array does not change.
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];
}
}