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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。

示例：

给定 nums = [-2, 0, 3, -5, 2, -1]，求和函数为 sumRange()

sumRange(0, 2) -> 1
sumRange(2, 5) -> -1
sumRange(0, 5) -> -3

说明:

1. 你可以假设数组不可变。
2. 会多次调用 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);
 */

...