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  • [LeetCode] Range Sum Query

    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:

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

    求一个数组的范围和,可以使用动态规划来计算。

    dp[x]数组表示原数组nums中前x个元素之和。如果要求nums中i~j内元素和,就要计算dp[j + 1] - dp[i]即可。

    class NumArray {
    public:
        NumArray(vector<int> nums) : dp(nums.size() + 1, 0) {
            for (int i = 1; i < dp.size(); i++)
                dp[i] = dp[i - 1] + nums[i - 1];
        }
        
        int sumRange(int i, int j) {
            return dp[j + 1] - dp[i];
        }
    private:
        vector<int> dp; 
    };
    // 29 ms
    /**
     * Your NumArray object will be instantiated and called as such:
     * NumArray obj = new NumArray(nums);
     * int param_1 = obj.sumRange(i,j);
     */
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  • 原文地址:https://www.cnblogs.com/immjc/p/7860559.html
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