Can you solve this real interview question? Range Sum Query - Mutable - Given an integer array nums, handle multiple queries of the following types: 1. Update the value of an element in nums. 2. Calculate the sum of the elements of nums between indice
Implement theNumArrayclass: NumArray(int[] nums)Initializes the object with the integer arraynums. int sumRange(int left, int right)Returns thesumof the elements ofnumsbetween indicesleftandrightinclusive(i.e.nums[left] + nums[left + 1] + ... + nums[right]). Example 1: Input["NumArr...
sumRange会被频繁调用。 给定的是左上角的点和右下角的点 思路: 提前求出一些小矩阵的和,具体来说Sum[i][j]表示从点(0,0)到点(i,j)(这里的点是指的左上角的点和右下角的点)所组成的矩阵的所有元素的和。 然后任意矩阵(row1,col1)->(row2,col2)的范围内元素的和为Sum(row2,col2) - Sum(ro...
}intsumRange(inti,intj) {if(i==0)returnsum[j+1];returnsum[j+1]-sum[i]; }private: vector<int>sum; };
LeetCode: 307. Range Sum Query - Mutable 题目描述 Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive. The update(i, val) function modifies nums by updating the element at index ...
https://leetcode.com/problems/range-sum-query-immutable/题目: 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 ...
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另外一种思路,在数组最前面加上一个0。这样一来,就有sumRange(i, j) = sum[j + 1] - sum[i],避免了可能的越界。 Code //Runtime: 169msclass NumArray{public:vector<int>sum;NumArray(vector<int>nums){inttmp=0;sum.push_back(tmp);intnumSize=nums.size();for(inti=0;i<numSize;i++){...
* LeetCode 307. Range Sum Query - Mutable * 区间求和,单点修改问题 * 其解决方法主要为树状数组和线段树 * * 此代码使用树状数组 * * 两点需要注意: * 1.树状数组下标从 1 开始,题目中数据数组从 0 开始 * 解决:index 整体 +1 即可 * 2.树状数组支持单点增加,此题为单点修改。
sums_[row1][col2 + 1] + sums_[row1][col1]; } private: vector<vector<int>> sums_; }; // Your NumMatrix object will be instantiated and called as such: // NumMatrix numMatrix(matrix); // numMatrix.sumRegion(0, 1, 2, 3); // numMatrix.sumRegion(1, 2, 3, 4);©...