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Leetcode 303 Range Sum Query Immutable

发表于 更新于 分类于 Valine：
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Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.

Implement the NumArray class:

  • NumArray(int[] nums) Initializes the object with the integer array nums.
  • int sumRange(int i, int j) Return the sum of the elements of the nums array in the range [i, j] inclusive (i.e., sum(nums[i], nums[i + 1], ... , nums[j]))

Example 1:

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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 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))

Constraints:

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

动态规划

储存一个数组accumulateSums, 保存从0到i的和. 计算的时候通过accumulateSums来返回结果

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class NumArray {
    int[] accumulateSums;
    public NumArray(int[] nums) {
        accumulateSums = new int[nums.length];
        int sum = 0;
        for(int i = 0; i != nums.length; ++i)
        {
            sum += nums[i];
            accumulateSums[i] = sum;
        }
    }
    
    public int sumRange(int i, int j) {
        if(i > j)
            throw new RuntimeException("indexException");
        else
        {
            if(i == 0)
                return accumulateSums[j];
            else
            {
                return accumulateSums[j] - accumulateSums[i - 1];
            }
        }
        
    }
}

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