303. 区域和检索 - 数组不可变

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))

思路

题解

class NumArray {
    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] = sums[i] + nums[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);
 */

Immutable 不可变

标签：检索,nums,NumArray,int,303,sums,数组,sumRange
来源： https://blog.csdn.net/weixin_41884690/article/details/114282393