Sunday, January 31, 2016

LeetCode 240 - Search a 2D Matrix II


https://leetcode.com/problems/search-a-2d-matrix-ii/
Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
  • Integers in each row are sorted in ascending from left to right.
  • Integers in each column are sorted in ascending from top to bottom.
For example,
Consider the following matrix:
[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]
Given target = 5, return true.
  • Time Complexity: O(m + n)
Given target = 20, return false.
https://leetcode.com/discuss/48852/my-concise-o-m-n-java-solution
public boolean searchMatrix(int[][] matrix, int target) { if(matrix == null || matrix.length < 1 || matrix[0].length <1) { return false; } int col = matrix[0].length-1; int row = 0; while(col >= 0 && row <= matrix.length-1) { if(target == matrix[row][col]) { return true; } else if(target < matrix[row][col]) { col--; } else if(target > matrix[row][col]) { row++; } } return false; }
 it's also OK to search from the bottom-left point. Just check the code below.
public static boolean searchMatrix(int[][] matrix, int target) { if (matrix == null || matrix.length < 1 || matrix[0].length < 1) return false; int col = 0; int row = matrix.length - 1; while (col <= matrix[0].length - 1 && row >= 0) { if (target == matrix[row][col]) return true; else if (target < matrix[row][col]) row--; else if (target > matrix[row][col]) col++; } return false; }
http://bookshadow.com/weblog/2015/07/23/leetcode-search-2d-matrix-ii/

O(m * logn)解法:循环枚举行,二分查找列

def searchMatrix(self, matrix, target): y = len(matrix[0]) - 1 def binSearch(nums, low, high): while low <= high: mid = (low + high) / 2 if nums[mid] > target: high = mid - 1 else: low = mid + 1 return high for x in range(len(matrix)): y = binSearch(matrix[x], 0, y) if matrix[x][y] == target: return True return False

O(n ^ 1.58)解法:

参考:https://leetcode.com/discuss/47528/c-with-o-m-n-complexity
分治法,以矩形中点为基准,将矩阵拆分成左上,左下,右上,右下四个区域
若中点值 < 目标值,则舍弃左上区域,从其余三个区域再行查找
若中点值 > 目标值,则舍弃右下区域,从其余三个区域再行查找
时间复杂度递推式:T(n) = 3T(n/2) + c
相关博文:http://articles.leetcode.com/2010/10/searching-2d-sorted-matrix-part-ii.html
public boolean searchMatrix(int[][] matrix, int target) { int n=matrix.length, m=matrix[0].length; return helper(matrix,0,n-1,0,m-1,target); } boolean helper(int[][] matrix, int rowStart, int rowEnd, int colStart, int colEnd, int target ){ if(rowStart>rowEnd||colStart>colEnd){ return false; } int rm=(rowStart+rowEnd)/2, cm=(colStart+colEnd)/2; if(matrix[rm][cm]== target){ return true; } else if(matrix[rm][cm] >target){ return helper(matrix, rowStart, rm-1,colStart, cm-1,target)|| helper(matrix, rm, rowEnd, colStart,cm-1,target) || helper(matrix, rowStart, rm-1,cm, colEnd,target); } else{ return helper(matrix, rm+1, rowEnd, cm+1,colEnd,target)|| helper(matrix, rm+1, rowEnd, colStart,cm,target) || helper(matrix, rowStart, rm,cm+1, colEnd,target); }

http://www.jyuan92.com/blog/leetcode-search-a-2d-matrix-ii/
Divide-Conquer,Based on the mid of the matrix, divide the whole matrix into four parts: left-top, left-bottom, right-top, right-bottom
  • If the mid is smaller than target, abandon the left-top area, recursion from the other three areas.
  • If the mid is larger than target, abandon the right-bottom area, recursion from the other three ares.
  • Time Complexity formula:T(n) = 3T(n/2) + c
public boolean searchMatrixImprove(int[][] matrix, int target) {
    if (null == matrix || matrix.length == 0) {
        return false;
    }
    return helper(matrix, 0, matrix.length - 1, 0, matrix[0].length - 1, target);
}
private boolean helper(int[][] matrix, int rowStart, int rowEnd, int colStart, int colEnd, int target) {
    if (rowStart > rowEnd || colStart > colEnd) {
        return false;
    }
    int rowMid = rowStart + (rowEnd - rowStart) / 2;
    int colMid = colStart + (colEnd - colStart) / 2;
    if (matrix[rowMid][colMid] == target) {
        return true;
    } else if (matrix[rowMid][colMid] > target) {
        return helper(matrix, rowStart, rowMid - 1, colStart, colMid - 1, target) ||
                helper(matrix, rowMid, rowEnd, colStart, colMid - 1, target) ||
                helper(matrix, rowStart, rowMid - 1, colMid, colEnd, target);
    } else {
        return helper(matrix, rowMid + 1, rowEnd, colMid + 1, colEnd, target) ||
                helper(matrix, rowMid + 1, rowEnd, colStart, colMid, target) ||
                helper(matrix, rowStart, rowMid, colMid + 1, colEnd, target);
    }
}
Searching a 2D Sorted Matrix II | LeetCode
From book - cracking code
Coordinate findElement(int[][] matrix, Coordinate or1g1n, Coordinate dest, int x){
 if (!origin.inbounds (matrix) || ! dest.inbounds(matrix)) {
 return null;
}
if (matrix[origin.row][origin.column]
return origin;
} else if (!origin.isBefore(dest)) {
return null;
 }

 /* Set start to start of diagonal and end to the end of the diagonal. Since the
 * grid may not be square, the end of the diagonal may not equal dest. */
 Coordinate start= (Coordinate) origin.clone();
 int diagDist = Math.min(dest.row - origin.row, dest.column - origin.column);
 Coordinate end= new Coordinate(start.row + diagDist, start.column + diagDist);
 Coordinate p = new Coordinate(0, 0);

 /* Do binary search on the diagonal, looking for the first element> x */
 while (start.isBefore(end)) {
 p.setToAverage(start, end);
 if (x > matrix[p.row][p.column]) {
 start.row = p.row + 1;
 start.column = p.column + 1;
 } else {
 end.row = p.row - 1;
 end.column = p.column - 1;
 }
 }

 /* Split the grid into quadrants. Search the bottom left and the top right. */
 return partitionAndSearch(matrix, origin, dest, start, x);
 }

 Coordinate partitionAndSearch(int[][] matrix, Coordinate origin, Coordinate dest,
 Coordinate pivot, int x) {
 Coordinate lowerLeftOrigin = new Coordinate(pivot.row, origin.column);
 Coordinate lowerLeftDest = new Coordinate(dest.row, pivot.column - 1);
 Coordinate upperRightOrigin = new Coordinate(origin.row, pivot.column);
 coordinate upperRightDest = new Coordinate(pivot.row - 1, dest.column);

 Coordinate lowerLeft = findElement(matrix, lowerLeftOrigin, lowerLeftDest, x);
 if (lowerleft == null) {
 return findElement(matrix, upperRightOrigin, upperRightDest, x);
 }
 return lowerleft;
 }

 Coordinate findElement(int[][] matrix, int x) {
 Coordinate origin = new Coordinate(0, 0);
 coordinate dest = new Coordinate(matrix.length - 1, matrix[0].length - 1);
 return findElement(matrix, origin, dest, x);
 }

 public class Coordinate implements Cloneable {
 public int row, column;
 public Coordinate(int r, int c) {
 row = r;
 column = c;
 }

 public boolean inbounds(int[][] matrix) {
 return row >= 0 && column >= 0 &&
 row< matrix.length && column< matrix[0].length;
 }

 public boolean isBefore(Coordinate p) {
 return row<= p.row && column<= p.column;
 }

 public Object clone() {
 return new Coordinate(row, column);
 }

 public void setToAverage(Coordinate min, Coordinate max) {
 row = (min.row + max.row) / 2;
 column = (min.column + max.column)/ 2;
 }
 }

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