力扣303. 区域和检索 - 数组不可变

Problem: 303. 区域和检索 - 数组不可变

题目描述

思路

创建前缀和数组preSum，其中preSum[i]处元素值为nums[0] - nums[i - 1]处元素值得和，当调用sumRange函数时直接返回preSum[right + 1] - preSum[left]

复杂度

函数sumRange的时间空间复杂度
时间复杂度:

$O(1)$

空间复杂度:

$O(1)$

Code

class NumArray {
    // Prefix array
    private int[] preSum;

    /**
     * Construct prefix sum
     *
     * @param nums Given array
     */
    public NumArray(int[] nums) {
        preSum = new int[nums.length + 1];
        for (int i = 1; i < preSum.length; ++i) {
            preSum[i] = preSum[i - 1] + nums[i - 1];
        }
    }

    /**
     * Query the sum of the closed interval [left, right]
     *
     * @param left  Left boundary
     * @param right Right boundary
     * @return int
     */
    public int sumRange(int left, int right) {
        return preSum[right + 1] - preSum[left];
    }
}

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