LeetCode 303. Range Sum Query - Immutable（数组求和）

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本文介绍了一种使用前缀和数组优化区间求和操作的方法，通过预先计算数组元素的累加和，使得在查询任意区间的和时能够实现高效的O(1)时间复杂度。

原题网址：https://leetcode.com/problems/range-sum-query-immutable/

Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.

Example:

Given nums = [-2, 0, 3, -5, 2, -1]

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

Note:

You may assume that the array does not change.
There are many calls to sumRange function.

方法：保存数组从0～i的累加值。

public class NumArray {
    private int[] sums;
    public NumArray(int[] nums) {
        sums = new int[nums.length];
        for(int i=0; i<nums.length; i++) {
            if (i==0) sums[i] = nums[i];
            else sums[i] = sums[i-1] + nums[i];
        }
    }

    public int sumRange(int i, int j) {
        if (i==0) return sums[j];
        return sums[j]-sums[i-1];
    }
}


// Your NumArray object will be instantiated and called as such:
// NumArray numArray = new NumArray(nums);
// numArray.sumRange(0, 1);
// numArray.sumRange(1, 2);