leetcode刷題筆記303題 區域和檢索 - 陣列不可變
阿新 • • 發佈:2020-11-25
leetcode刷題筆記303題 區域和檢索 - 陣列不可變
問題描述:
給定一個整數陣列 nums,求出陣列從索引 i 到 j(i ≤ j)範圍內元素的總和,包含 i、j 兩點。
實現 NumArray 類:
NumArray(int[] nums) 使用陣列 nums 初始化物件
int sumRange(int i, int j) 返回陣列 nums 從索引 i 到 j(i ≤ j)範圍內元素的總和,包含 i、j 兩點(也就是 sum(nums[i], nums[i + 1], ... , nums[j]))示例:
輸入:
["NumArray", "sumRange", "sumRange", "sumRange"]
[[[-2, 0, 3, -5, 2, -1]], [0, 2], [2, 5], [0, 5]]
輸出:
[null, 1, -1, -3]解釋:
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
最多呼叫 104 次 sumRange 方法
//一維字首和處理 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);
*/