308. Range Sum Query 2D

题目：

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) = (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) -> 8
update(3, 2, 2)
sumRegion(2, 1, 4, 3) -> 10

Note:

  1. The matrix is only modifiable by the update function.
  2. You may assume the number of calls to update and sumRegion function is distributed evenly.
  3. You may assume that row1 ≤ row2 and col1 ≤ col2.

链接： http://leetcode.com/problems/range-sum-query-2d-mutable/

题解：

二维Range Sum Query mutable，我了个去，第一次Leetcode写了超过140行代码...足够臭长了吧，但是居然能ac，还是很高兴 -____-!! 原理是要构建一个2D Segment Tree或者 2D Fenwick Tree。由于上一题是先做的Segment Tree，这回也先写2D Segment Tree。构建2D Segment Tree依然是使用Divide and Conquer，我们要把整个平面分成4个部分，所以2D Segment Tree也是一个Quad Tree，每个节点有四个子节点，NW, NE, SW, SE， 节点的sum是四个子节点的sum。这样我们就可以用与1D Segment Tree类似的方法来写rangeSum以及update。要注意rangeSum时的判断，有好几种情况，比较复杂。

2D Segment Tree: Time Complexity - O(mn) build，O(logmn) update， O(logmn) rangeSum ， Space Complexity - O(mn)  复杂度算得不是很清楚，很可能不正确，二刷再继续改正。

public class NumMatrix {
    private class SegmentTreeNode2D {
        public int tlRow;
        public int tlCol;
        public int brRow;
        public int brCol;
        public int sum;
        public SegmentTreeNode2D nw, ne, sw, se;
        
        public SegmentTreeNode2D(int tlRow, int tlCol, int brRow, int brCol) {
            this.tlRow = tlRow;
            this.tlCol = tlCol;
            this.brRow = brRow;
            this.brCol = brCol;
            this.sum = 0;              
        }
    }
    
    public SegmentTreeNode2D root;
    
    public NumMatrix(int[][] matrix) {
        if(matrix == null || matrix.length == 0) {
            return;
        }
        root = buildTree(matrix, 0, 0, matrix.length - 1, matrix[0].length - 1);    
    }

    public void update(int row, int col, int val) {
        update(root, row, col, val);
    }
    
    private void update(SegmentTreeNode2D node, int row, int col, int val) {
        if(node.tlRow == row && node.brRow == row && node.tlCol == col && node.brCol == col) {
            node.sum = val;
            return;
        }
        int rowMid = node.tlRow + (node.brRow - node.tlRow) / 2;
        int colMid = node.tlCol + (node.brCol - node.tlCol) / 2;
        if(row <= rowMid) {
            if(col <= colMid) {
                update(node.nw, row, col, val);
            } else {
                update(node.ne, row, col, val);
            }
        } else {
            if(col <= colMid) {
                update(node.sw, row, col, val);
            } else {
                update(node.se, row, col, val);
            }
        }
        
        node.sum = 0;
        if(node.nw != null) {
            node.sum += node.nw.sum;
        }
        if(node.ne != null) {
            node.sum += node.ne.sum;
        }
        if(node.sw != null) {
            node.sum += node.sw.sum;
        }
        if(node.se != null) {
            node.sum += node.se.sum;
        }
    }

    
    public int sumRegion(int row1, int col1, int row2, int col2) {
        return sumRegion(root, row1, col1, row2, col2);    
    }
    
    private int sumRegion(SegmentTreeNode2D node, int tlRow, int tlCol, int brRow, int brCol) {
        if(node.tlRow == tlRow && node.tlCol == tlCol && node.brRow == brRow && node.brCol == brCol) {
            return node.sum;
        }
        int rowMid = node.tlRow + (node.brRow - node.tlRow) / 2;
        int colMid = node.tlCol + (node.brCol - node.tlCol) / 2;
        if(brRow <= rowMid) {  // top-half plane
            if(brCol <= colMid) {         // north-west quadrant
                return sumRegion(node.nw, tlRow, tlCol, brRow, brCol);
            } else if(tlCol > colMid) {    // north-east quadrant 
                return sumRegion(node.ne, tlRow, tlCol, brRow, brCol);
            } else {                // intersection between nw and ne
                return sumRegion(node.nw, tlRow, tlCol, brRow, colMid) + sumRegion(node.ne, tlRow, colMid + 1, brRow, brCol);
            }
        } else if(tlRow > rowMid) {         // bot-half plane
            if(brCol <= colMid) {         // south-west quadrant
                return sumRegion(node.sw, tlRow, tlCol, brRow, brCol);
            } else if(tlCol > colMid) {    // south-east quadrant 
                return sumRegion(node.se, tlRow, tlCol, brRow, brCol);
            } else {                //intersection between sw and sw
                return sumRegion(node.sw, tlRow, tlCol, brRow, colMid) + sumRegion(node.se, tlRow, colMid + 1, brRow, brCol);                
            }
        } else {                // full-plane intersection
            if(brCol <= colMid) {         // left half plane
                return sumRegion(node.nw, tlRow, tlCol, rowMid, brCol) + sumRegion(node.sw, rowMid + 1, tlCol, brRow, brCol) ;
            } else if(tlCol > colMid) {    // right half plane 
                return sumRegion(node.ne, tlRow, tlCol, rowMid, brCol) + sumRegion(node.se, rowMid + 1, tlCol, brRow, brCol) ;
            } else {                // full-plane intersection
                return sumRegion(node.nw, tlRow, tlCol, rowMid, colMid)
                     + sumRegion(node.ne, tlRow, colMid + 1, rowMid, brCol)
                     + sumRegion(node.sw, rowMid + 1, tlCol, brRow, colMid)
                     + sumRegion(node.se, rowMid + 1, colMid + 1, brRow, brCol);
            }
        }
    }
    
    
    private SegmentTreeNode2D buildTree(int[][] matrix, int tlRow, int tlCol, int brRow, int brCol) {        
        if(tlRow > brRow || tlCol > brCol) {
            return null;
        } else {
            SegmentTreeNode2D node = new SegmentTreeNode2D(tlRow, tlCol, brRow, brCol);
            if(tlRow == brRow && tlCol == brCol) {
                node.sum = matrix[tlRow][tlCol];
            } else {
                int rowMid = tlRow + (brRow - tlRow) / 2;
                int colMid = tlCol + (brCol - tlCol) / 2;
                node.nw = buildTree(matrix, tlRow, tlCol, rowMid, colMid); 
                node.ne = buildTree(matrix, tlRow, colMid + 1, rowMid, brCol);
                node.sw = buildTree(matrix, rowMid + 1, tlCol, brRow, colMid);
                node.se = buildTree(matrix, rowMid + 1, colMid + 1, brRow, brCol);
                node.sum = 0;
                if(node.nw != null) {
                    node.sum += node.nw.sum;
                }
                if(node.ne != null) {
                    node.sum += node.ne.sum;
                }
                if(node.sw != null) {
                    node.sum += node.sw.sum;
                }
                if(node.se != null) {
                    node.sum += node.se.sum;
                }                
            }
            return node;
        }
    }
}


// Your NumMatrix object will be instantiated and called as such:
// NumMatrix numMatrix = new NumMatrix(matrix);
// numMatrix.sumRegion(0, 1, 2, 3);
// numMatrix.update(1, 1, 10);
// numMatrix.sumRegion(1, 2, 3, 4);

2D Fenwick Tree:  -- 看了Quora一个acm大神的post以后，我决定还是要使用2D Fenwick Tree来做这题。 “https://www.quora.com/How-does-a-2D-segment-tree-work” ，代码肯定比Segment Tree简洁，而且速度也会更快。基本方法和1D非常类似，这种方法甚至可以简单地扩展到更高维度。

Time Complexity - O(mnlogm * logn) build,  O(logmn) update， O(logmn) rangeSum， Space Complexity - O(mn)

public class NumMatrix {
    private int BIT2D[][];
    private int matrix[][];
    
    public NumMatrix(int[][] matrix) {
        if(matrix == null || matrix.length == 0) {
            return;
        }
        BIT2D = new int[matrix.length + 1][matrix[0].length + 1];
        this.matrix = new int[matrix.length][matrix[0].length];
        for(int i = 0; i < matrix.length; i++) {
            for(int j = 0; j < matrix[0].length; j++) {
                update(i, j, matrix[i][j]);
            }
        }
    }

    public void update(int row, int col, int val) {
        int delta = val - matrix[row][col];
        matrix[row][col] = val;
        for(int i = row + 1; i < BIT2D.length; i += i & (-i)) {         //also equals to i |= i + 1
            for(int j = col + 1; j < BIT2D[0].length; j += j & (-j)) {
                BIT2D[i][j] += delta;
            }
        }
    }
    

    public int sumRegion(int row1, int col1, int row2, int col2) {
        return getSum(row2 + 1, col2 + 1) - getSum(row1, col2 + 1) - getSum(row2 + 1, col1) + getSum(row1, col1); 
    }
    
    private int getSum(int row, int col) {
        int sum = 0;
        for(int i = row; i > 0; i -= i & (-i)) {
            for(int j = col; j > 0; j -= j & (-j)) {
                sum += BIT2D[i][j];
            }
        }
        return sum;
    }
}


// Your NumMatrix object will be instantiated and called as such:
// NumMatrix numMatrix = new NumMatrix(matrix);
// numMatrix.sumRegion(0, 1, 2, 3);
// numMatrix.update(1, 1, 10);
// numMatrix.sumRegion(1, 2, 3, 4);

Reference:

https://stackoverflow.com/questions/25121878/2d-segment-quad-tree-explanation-with-c/25122078#25122078

https://sites.google.com/site/indy256/algo/fenwick_tree_2d
http://www.hawstein.com/posts/binary-indexed-trees.html
https://www.topcoder.com/community/data-science/data-science-tutorials/binary-indexed-trees/
http://www.wohenniu.com/thread-872-1-1.html
http://bookshadow.com/leetcode/
http://cs.nyu.edu/courses/spring14/CSCI-UA.0480-004/
https://web.stanford.edu/class/cs97si/03-data-structures.pdf
http://stackoverflow.com/questions/9452701/ukkonens-suffix-tree-algorithm-in-plain-english

https://leetcode.com/discuss/71025/segmentation-tree-736ms-indexed-tree-492ms-based-solutions

https://leetcode.com/discuss/70992/c-solution-using-2d-binary-index-tree-easy-to-understand

https://leetcode.com/discuss/72685/share-my-java-2-d-binary-indexed-tree-solution

https://leetcode.com/discuss/71046/java-2d-binary-indexed-tree-solution-80ms

https://leetcode.com/discuss/70948/15ms-easy-to-understand-java-solution

https://leetcode.com/discuss/71169/java-2d-binary-indexed-tree-solution-clean-and-short-17ms

https://leetcode.com/problems/range-sum-query-2d-mutable/

http://www.lxway.com/5152462.htm

https://www.topcoder.com/community/data-science/data-science-tutorials/binary-indexed-trees/

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原文地址：https://www.cnblogs.com/yrbbest/p/5058571.html