(LeetCode74) Search a 2D Matrix

Write an efficient algorithm, searches for a value in a m x n Matrix. This matrix has the following properties:

• Integers in each row is sorted from the left to the right.
• The first integer of each row was greater than the last integer of the previous row.

For example,

Consider the following matrix:

[  [1,   3,  5,  7],  [Ten, One,],  [23, 30, 34, 50]]

Given Target = 3 , return true .

Topic:

In a m*n two-dimensional array, each row is incremented from left to right, and the first element of each row is larger than the last element in the previous row.

Determines whether an element is in the array.

Ideas:

One implication of the problem is that each column is incremented from top to bottom.

Method 1:

If you expand a two-dimensional array into rows, it is a sorted one-dimensional array, so it is easy to find the answer by looking at the binary of a one-dimensional array.

Method 2:

Given the regularity of the array, select the upper-right number of the array look-up range, if it is equal to the number found, returns true if the number is larger than the lookup, and if it is smaller than the lookup number, the row of the number is excluded from the lookup range.

Method 3:

Finds the row of the element first by two points, and then finds the element by two points in the same row.

Code:

1, one-dimensional array method:

classSolution { Public:    BOOLSearchmatrix (vector<vector<int> > &matrix,inttarget) {        intRows=matrix.size (); intcols=matrix[0].size (); intleft=0, right= (rows*cols-1); intMid,r,c,val;  while(left<=Right ) {Mid=left+ (Right-left) >>1); R=mid/cols; C=mid%cols; if(matrix[r][c]==target)return true; if(matrix[r][c]<target) left=mid+1; Else Right=mid-1; }        return false; }};

2. Upper right corner element comparison method

classSolution { Public:    BOOLSearchmatrix (vector<vector<int> > &matrix,inttarget) {        intRows=matrix.size (); intcols=matrix[0].size (); intR=0, c=cols-1;  while(R<rows && c>=0){            if(target==Matrix[r][c])return true; if(target<Matrix[r][c]) C--; ElseR++; }        return false; }};

3, two-point lookup line method

classSolution { Public:    BOOLSearchmatrix (vector<vector<int> > &matrix,inttarget) {        intRows=matrix.size (); intcols=matrix[0].size (); intlrow=0, rrow=rows-1; intMidrow;  while(lrow<=Rrow) {Midrow=lrow+ (Rrow-lrow) >>1); if(matrix[midrow][0]==target | | matrix[midrow][cols-1]==target)return true; if(matrix[midrow][0]<target && matrix[midrow][cols-1]>target) Break; if(matrix[midrow][0]>target) Rrow=midrow-1; ElseLrow=midrow+1; }                if(lrow<=Rrow) {            intLcol=0, rcol=cols-1; intMidcol;  while(lcol<=Rcol) {Midcol=lcol+ (Rcol-lcol) >>1); if(matrix[midrow][midcol]==target)return true; if(matrix[midrow][midcol]>target) Rcol=midcol-1; ElseLCol=midcol+1; }        }                return false; }};

(LeetCode74) Search a 2D Matrix