由于我们之前做过一维数组的可变和不可变的情况Range Sum Query - Mutable和Range Sum Query - Immutable,那么为了能够通过OJ,我们还是需要用到树状数组Binary Indexed Tree(参见Range Sum Query - Mutable),其查询和修改的复杂度均为O(logn),那么我们还是要建立树状数组,我们根据数组中的...
Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2). Range Sum Query 2D The above rectangle (with the red border) is defined by (row1, col1)= (2, 1) and (row2, col2)...
先贴一个prefix Sum的做法 时间复杂度 O(N) update, O(N) query class NumMatrix { int[][] matrix; int[][] prefixSum; int N, M; public NumMatrix(int[][] matrix) { N = matrix.length; if (N == 0) return; M = matrix[0].length; this.matrix = new int[N][M]; for (int i...
sum=m[iy0][ix0];return; }intxmid =getMidX();intymid =getMidY(); ul=newNode(m, ix0, iy0, xmid, ymid); sum+= ul->sum;if(ix1 >xmid) { ur=newNode(m, xmid +1, iy0, ix1, ymid); sum+= ur->sum; }if(iy1 >ymid) { dl=newNode(m, ix0, ymid +1, xmid, i...
Range Sum Query 2D - Mutable Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2). The above rectangle (with the red border) is defined by (row1, col1) = (2, 1) and (...
link:https://www.lintcode.com/problem/range-sum-query-2d-mutable/description Description Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2). ...
Range Sum Query 2D - Mutable Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2). The above rectangle (with the red border) is defined by (row1, col1) = (2, 1) and (...
(int i, int val) { nodes[i += N] = val; while (i > 0) { nodes[i / 2] = nodes[i] + nodes[i ^ 1]; i /= 2; } } int sumRange(int i, int j) { int sum = 0; for (i += N, j += N; i <= j; i /= 2, j /= 2) { // 左边界为右结点, 加上自己,并...
(2*root+1,left,mid,nums);buildTree(2*root+2,mid+1,right,nums);tree[root]=tree[2*root+1]+tree[2*root+2];// 根节点的和,为左右子树的和}intquery(inti,intj){returnquery(0,i,j,0,N-1);}intquery(introot,intleft,intright,inttreeLeft,inttreeRight){// 在以root为根的树中,查询...
Range Sum Query - Mutable 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 i to val. Example: Given nums = [1, 3, 5] ...