LeetCode 303. Range Sum Query - Immutable

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最新推荐文章于 2025-05-24 17:38:15 发布

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博客围绕整数数组的区间和查询问题展开。常规做法在query时每次相加，但因构造次数少、query次数多，需优化query时间复杂度。可在构造时存放当前index前数字和，query时相减，若left为0则直接返回right值，并给出了Java实现的运行时间和内存使用情况。

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Given an integer array nums, handle multiple queries of the following type:

Calculate the sum of the elements of nums between indices left and right inclusive where left <= right.

Implement the NumArray class:

NumArray(int[] nums) Initializes the object with the integer array nums.
int sumRange(int left, int right) Returns the sum of the elements of nums between indices left and right inclusive (i.e. nums[left] + nums[left + 1] + ... + nums[right]).

Example 1:

Input
["NumArray", "sumRange", "sumRange", "sumRange"]
[[[-2, 0, 3, -5, 2, -1]], [0, 2], [2, 5], [0, 5]]
Output
[null, 1, -1, -3]

Explanation
NumArray numArray = new NumArray([-2, 0, 3, -5, 2, -1]);
numArray.sumRange(0, 2); // return (-2) + 0 + 3 = 1
numArray.sumRange(2, 5); // return 3 + (-5) + 2 + (-1) = -1
numArray.sumRange(0, 5); // return (-2) + 0 + 3 + (-5) + 2 + (-1) = -3

Constraints:

1 <= nums.length <= 104
-105 <= nums[i] <= 105
0 <= left <= right < nums.length
At most 104 calls will be made to sumRange.

Accepted

397,212

Submissions

如果按照正常的想法很简单，就是在query的时候每次加一遍。但是因为构造的次数少，query的次数多，所以需要optimize query的时间复杂度，达到O(n)构造，O(1) query。那么构造的时候就可以存放当前index之前的数字和，query的时候只需要二者相减就行。但是需要注意的是，query时两个相减，如果left是0的话，应该直接返回right的值。

Runtime: 21 ms, faster than 49.58% of Java online submissions for Range Sum Query - Immutable.

Memory Usage: 48.8 MB, less than 65.91% of Java online submissions for Range Sum Query - Immutable.

class NumArray {

    int[] sums;
    public NumArray(int[] nums) {
        sums = new int[nums.length];
        sums[0] = nums[0];
        for (int i = 1; i < nums.length; i++) {
            sums[i] = sums[i - 1] + nums[i];
        }
    }
    
    public int sumRange(int left, int right) {
        if (left == 0) {
            return sums[right];
        }
        return sums[right] - sums[left - 1];
    }
}

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