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303. 区域和检索 - 数组不可变

English Version

题目描述

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

解法

Python3

class NumArray:

    def __init__(self, nums: List[int]):
        n = len(nums)
        self.sums = [0] * (n + 1)
        for i in range(n):
            self.sums[i + 1] = nums[i] + self.sums[i]


    def sumRange(self, i: int, j: int) -> int:
        return self.sums[j + 1] - self.sums[i]


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

Java

class NumArray {

    private int[] sums;

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

    public int sumRange(int i, int j) {
        return sums[j + 1] - sums[i];
    }
}

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

...