leetcode刷题笔记303题 区域和检索 - 数组不可变
问题描述：
给定一个整数数组 nums，求出数组从索引 i 到 j（i ≤ j）范围内元素的总和，包含 i、j 两点。
实现 NumArray 类：
NumArray(int[] nums) 使用数组 nums 初始化对象
int sumRange(int i, int j) 返回数组 nums 从索引 i 到 j（i ≤ j）范围内元素的总和，包含 i、j 两点（也就是 sum(nums[i], nums[i + 1], ... , nums[j])）示例：
输入：
["NumArray", "sumRange", "sumRange", "sumRange"]
[[[-2, 0, 3, -5, 2, -1]], [0, 2], [2, 5], [0, 5]]
输出：
[null, 1, -1, -3]解释：
NumArray numArray = new NumArray([-2, 0, 3, -5, 2, -1]);
numArray.sumRange(0, 2); // return 1 ((-2) + 0 + 3)
numArray.sumRange(2, 5); // return -1 (3 + (-5) + 2 + (-1))
numArray.sumRange(0, 5); // return -3 ((-2) + 0 + 3 + (-5) + 2 + (-1))提示：
0 <= nums.length <= 104
-105 <= nums[i] <= 105
0 <= i <= j < nums.length
最多调用 104 次 sumRange 方法
//一维前缀和处理
class NumArray(nums: Array[Int]) {
val arr = Array.fill(nums.length+1)(0)
for (i <- 1 to nums.length) {
arr(i) = arr(i-1) + nums(i-1)
}
def sumRange(i: Int, j: Int): Int = {
return arr(j+1) - arr(i)
}
}
/**
* Your NumArray object will be instantiated and called as such:
* var obj = new NumArray(nums)
* var param_1 = obj.sumRange(i,j)
*/
type NumArray struct {
arr []int
}
func Constructor(nums []int) NumArray {
nArr := make([]int, len(nums)+1)
for i := 1; i <= len(nums); i++ {
nArr[i] = nArr[i-1] + nums[i-1]
}
return NumArray{nArr}
}
func (this *NumArray) SumRange(i int, j int) int {
return this.arr[j+1] - this.arr[i]
}
/**
* Your NumArray object will be instantiated and called as such:
* obj := Constructor(nums);
* param_1 := obj.SumRange(i,j);
*/