[LeetCode308]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).Range Sum Query 2DThe above rectangle (with the red border) is defined by (row1, col1) = (2, 1) and (row2, col2) = (4, 3), which contains sum = 8.Example:Given matrix = [      [3, 0, 1, 4, 2],      [5, 6, 3, 2, 1],      [1, 2, 0, 1, 5],      [4, 1, 0, 1, 7],      [1, 0, 3, 0, 5]    ]sumRegion(2, 1, 4, 3) -> 8update(3, 2, 2)sumRegion(2, 1, 4, 3) -> 10Note:    The matrix is only modifiable by the update function.    You may assume the number of calls to update and sumRegion function is distributed evenly.    You may assume that row1 ≤ row2 and col1 ≤ col2.Hide Tags Segment Tree Binary Indexed TreeHide Similar Problems (M) Range Sum Query 2D - Immutable (M) Range Sum Query - Mutable

这道题看着超级花骚，然后code也会很长，但其实就是segment tree的变形，1D的懂了就可以慢慢摸索2D的。。。

class SegmentTreeSumNode {public:    int sum;    SegmentTreeSumNode *negibor[4] = {NULL, NULL, NULL, NULL};    pair<int, int> leftTop = make_pair(0, 0);    pair<int, int> rightBottom = make_pair(0, 0);    SegmentTreeSumNode(int sum) : sum(sum){}};class NumMatrix {    SegmentTreeSumNode* root;    vector<vector<int>> nums;public:    NumMatrix(vector<vector<int>> &matrix) {        if(matrix.empty()) return;        nums = matrix;        int m = matrix.size(), n = matrix[0].size();        if(!m || !n) return;        root = buildSegmentTree(matrix, make_pair(0,0), make_pair(m-1, n-1));    }    void update(int row, int col, int val) {        int diff = val - nums[row][col];        if(!diff) return;        nums[row][col] = val;        update(row, col, diff, root);    }    int sumRegion(int row1, int col1, int row2, int col2) {        int res = 0;        if(root) sumRange(row1, col1, row2, col2, root, res);        return res;    }    SegmentTreeSumNode* buildSegmentTree(vector<vector<int>>& matrix, pair<int, int> start, pair<int , int> end){        if(start.first > end.first || start.second > end.second) return NULL;        SegmentTreeSumNode* node = new SegmentTreeSumNode(0);        node->leftTop = start;        node->rightBottom = end;        if(start == end){            node->sum = matrix[start.first][start.second];            return node;        }        int midx = (start.first + end.first)/2;        int midy = (start.second + end.second)/2;        node->negibor[0] = buildSegmentTree(matrix, start, make_pair(midx, midy));        node->negibor[1] = buildSegmentTree(matrix, make_pair(start.first, midy+1), make_pair(midx, end.second));        node->negibor[2] = buildSegmentTree(matrix, make_pair(midx + 1, start.second), make_pair(end.first, midy));        node->negibor[3] = buildSegmentTree(matrix, make_pair(midx+1, midy+1), end);        for (int i = 0; i<4; ++i) {            if(node->negibor[i]) node->sum += node->negibor[i]->sum;        }        return node;    }    void update(int row, int col, int diff, SegmentTreeSumNode* node){        if(row >= (node->leftTop).first && row <= (node->rightBottom).first && col >= (node->leftTop).second && col <= (node->rightBottom).second){            node->sum += diff;            for (int i = 0; i<4; ++i) {                if(node->negibor[i]) update(row, col, diff, node->negibor[i]);            }        }    }    void sumRange(int row1, int col1, int row2, int col2, SegmentTreeSumNode* node, int& res){        pair<int, int> start = node->leftTop;        pair<int, int> end = node->rightBottom;        int top = max(start.first, row1);        int bottom = min(end.first, row2);        if(bottom < top) return;        int left = max(start.second, col1);        int right = min(end.second, col2);        if(left > right) return;        if(row1 <= start.first && col1 <= start.second && row2 >= end.first && col2 >= end.second){            res += node->sum;            return;        }        for (int i = 0; i<4; ++i) {            if(node->negibor[i]) sumRange(row1, col1, row2, col2, node->negibor[i], res);        }    }};// Your NumMatrix object will be instantiated and called as such:// NumMatrix numMatrix(matrix);// numMatrix.sumRegion(0, 1, 2, 3);// numMatrix.update(1, 1, 10);// numMatrix.sumRegion(1, 2, 3, 4);

12/15/2015:update:

今天看了Fenwick tree, we can solve the problem with O(logmlogn):

class NumMatrix {public:    NumMatrix(vector<vector<int>> &matrix) {        if (matrix.size() == 0 || matrix[0].size() == 0) return;        nrow = matrix.size();        ncol = matrix[0].size();        nums = matrix;        BIT = vector<vector<int>> (nrow+1, vector<int>(ncol+1, 0));        for (int i = 0; i < nrow; i++)            for (int j = 0; j < ncol; j++)                add(i, j, matrix[i][j]);    }    void update(int row, int col, int val) {        int diff = val - nums[row][col];        add(row, col,diff);        nums[row][col] = val;    }    int sumRegion(int row1, int col1, int row2, int col2) {        int regionL = 0, regionS = 0;        int regionLeft = 0, regionTop = 0;        regionL = region(row2, col2);        if (row1 > 0 && col1 > 0) regionS = region(row1-1, col1-1);        if (row1 > 0) regionTop  = region(row1-1, col2);        if (col1 > 0) regionLeft = region(row2, col1-1);               return regionL - regionTop - regionLeft + regionS;    }private:    vector<vector<int>> nums;    vector<vector<int>> BIT;    int nrow = 0;    int ncol = 0;    void add(int row, int col, int val) {        row++;        col++;        while(row <= nrow) {            int colIdx = col;            while(colIdx <= ncol) {                BIT[row][colIdx] += val;                colIdx += (colIdx & (-colIdx));            }            row +=  (row & (-row));        }    }    int region(int row, int col) {        row++;        col++;        int res = 0;        while(row > 0) {            int colIdx = col;            while(colIdx > 0) {                res += BIT[row][colIdx];                colIdx -= (colIdx & (-colIdx));            }            row -= (row & (-row));        }        return res;    }};

youtube有个视频讲的蛮好，是1d的。。2d再研究一下吧。
这个可以看一下。。
https://leetcode.com/discuss/71169/java-2d-binary-indexed-tree-solution-clean-and-short-17ms