303. 区域和检索 - 数组不可变

题目描述

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] )

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

解法

方法一：前缀和

我们创建一个长度为 $n + 1$ 的前缀和数组 $s$，其中 $s[i]$ 表示前 $i$ 个元素的前缀和，即 $s[i] = \sum_{j=0}^{i-1} nums[j]$，那么索引 $[left, right]$ 之间的元素的和就可以表示为 $s[right + 1] - s[left]$。

初始化前缀和数组 $s$ 的时间复杂度为 $O(n)$，查询的时间复杂度为 $O(1)$。空间复杂度 $O(n)$。

Python3JavaC++GoTypeScriptRustJavaScriptPHPC

class NumArray:
    def __init__(self, nums: List[int]):
        self.s = list(accumulate(nums, initial=0))

    def sumRange(self, left: int, right: int) -> int:
        return self.s[right + 1] - self.s[left]


# Your NumArray object will be instantiated and called as such:
# obj = NumArray(nums)
# param_1 = obj.sumRange(left,right)

class NumArray {
    private int[] s;

    public NumArray(int[] nums) {
        int n = nums.length;
        s = new int[n + 1];
        for (int i = 0; i < n; ++i) {
            s[i + 1] = s[i] + nums[i];
        }
    }

    public int sumRange(int left, int right) {
        return s[right + 1] - s[left];
    }
}

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

class NumArray {
public:
    NumArray(vector<int>& nums) {
        int n = nums.size();
        s.resize(n + 1);
        for (int i = 0; i < n; ++i) {
            s[i + 1] = s[i] + nums[i];
        }
    }

    int sumRange(int left, int right) {
        return s[right + 1] - s[left];
    }

private:
    vector<int> s;
};

/**
 * Your NumArray object will be instantiated and called as such:
 * NumArray* obj = new NumArray(nums);
 * int param_1 = obj->sumRange(left,right);
 */

type NumArray struct {
    s []int
}

func Constructor(nums []int) NumArray {
    n := len(nums)
    s := make([]int, n+1)
    for i, v := range nums {
        s[i+1] = s[i] + v
    }
    return NumArray{s}
}

func (this *NumArray) SumRange(left int, right int) int {
    return this.s[right+1] - this.s[left]
}

/**
 * Your NumArray object will be instantiated and called as such:
 * obj := Constructor(nums);
 * param_1 := obj.SumRange(left,right);
 */

class NumArray {
    private s: number[];

    constructor(nums: number[]) {
        const n = nums.length;
        this.s = Array(n + 1).fill(0);
        for (let i = 0; i < n; ++i) {
            this.s[i + 1] = this.s[i] + nums[i];
        }
    }

    sumRange(left: number, right: number): number {
        return this.s[right + 1] - this.s[left];
    }
}

/**
 * Your NumArray object will be instantiated and called as such:
 * var obj = new NumArray(nums)
 * var param_1 = obj.sumRange(left,right)
 */

struct NumArray {
    s: Vec<i32>,
}

/**
 * `&self` means the method takes an immutable reference.
 * If you need a mutable reference, change it to `&mut self` instead.
 */
impl NumArray {
    fn new(mut nums: Vec<i32>) -> Self {
        let n = nums.len();
        let mut s = vec![0; n + 1];
        for i in 0..n {
            s[i + 1] = s[i] + nums[i];
        }
        Self { s }
    }

    fn sum_range(&self, left: i32, right: i32) -> i32 {
        self.s[(right + 1) as usize] - self.s[left as usize]
    }
}/**
 * Your NumArray object will be instantiated and called as such:
 * let obj = NumArray::new(nums);
 * let ret_1: i32 = obj.sum_range(left, right);
 */

/**
 * @param {number[]} nums
 */
var NumArray = function (nums) {
    const n = nums.length;
    this.s = Array(n + 1).fill(0);
    for (let i = 0; i < n; ++i) {
        this.s[i + 1] = this.s[i] + nums[i];
    }
};

/**
 * @param {number} left
 * @param {number} right
 * @return {number}
 */
NumArray.prototype.sumRange = function (left, right) {
    return this.s[right + 1] - this.s[left];
};

/**
 * Your NumArray object will be instantiated and called as such:
 * var obj = new NumArray(nums)
 * var param_1 = obj.sumRange(left,right)
 */

class NumArray {
    /**
     * @param Integer[] $nums
     */
    function __construct($nums) {
        $this->s = [0];
        foreach ($nums as $x) {
            $this->s[] = $this->s[count($this->s) - 1] + $x;
        }
    }

    /**
     * @param Integer $left
     * @param Integer $right
     * @return Integer
     */
    function sumRange($left, $right) {
        return $this->s[$right + 1] - $this->s[$left];
    }
}

/**
 * Your NumArray object will be instantiated and called as such:
 * $obj = NumArray($nums);
 * $ret_1 = $obj->sumRange($left, $right);
 */

typedef struct {
    int* s;
} NumArray;

NumArray* numArrayCreate(int* nums, int n) {
    int* s = malloc(sizeof(int) * (n + 1));
    s[0] = 0;
    for (int i = 0; i < n; i++) {
        s[i + 1] = s[i] + nums[i];
    }
    NumArray* obj = malloc(sizeof(NumArray));
    obj->s = s;
    return obj;
}

int numArraySumRange(NumArray* obj, int left, int right) {
    return obj->s[right + 1] - obj->s[left];
}

void numArrayFree(NumArray* obj) {
    free(obj->s);
    free(obj);
}

/**
 * Your NumArray struct will be instantiated and called as such:
 * NumArray* obj = numArrayCreate(nums, numsSize);
 * int param_1 = numArraySumRange(obj, left, right);

 * numArrayFree(obj);
*/

303. 区域和检索 - 数组不可变

题目描述

解法

方法一：前缀和

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