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303. 区域和检索 - 数组不可变
给定一个整数数组 nums,处理以下类型的多个查询:
计算索引 left 和 right (包含 left 和 right)之间的 nums 元素的 和 ,其中 left <= right
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 中索引 left 和 right 之间的元素的 总和 ,包含 left 和 right 两点(也就是 nums[left] + nums[left + 1] + ... + nums[right] )
javascript
示例 1:
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:
1 <= nums.length <= 104
-105 <= nums[i] <= 105
0 <= i <= j < nums.length
最多调用 104 次 sumRange 方法
- 来源:力扣(LeetCode)
- 链接:https://leetcode.cn/problems/range-sum-query-immutable
javascript
/**
* @param {number[]} nums
*/
var NumArray = function (nums) {
this.nums = [...nums]
}
/**
* @param {number} left
* @param {number} right
* @return {number}
*/
NumArray.prototype.sumRange = function (left, right) {
return this.nums.reduce((res, cur, index) => {
if (index >= left && index <= right) {
res += cur
}
return res
}, 0)
}
/**
* Your NumArray object will be instantiated and called as such:
* var obj = new NumArray(nums)
* var param_1 = obj.sumRange(left,right)
*/
var numArray = new NumArray([-2, 0, 3, -5, 2, -1])
console.log(numArray.sumRange(0, 2)) // return 1 ((-2) + 0 + 3)
console.log(numArray.sumRange(2, 5)) // return -1 (3 + (-5) + 2 + (-1))
console.log(numArray.sumRange(0, 5)) // return -3 ((-2) + 0 + 3 + (-5) + 2 + (-1))