Leetcode Search a 2D Matrix-----java

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 .

The topic is to give a matrix, according to a small arrival, and then given a number, determine whether the number in the matrix.

Is the application of the dichotomy method. Two methods, the first kind is slightly slower.

The first is to use the two-way, first find the number of trips, and then in this line to find the specific existence.

The second is the direct application of the dichotomy, more efficient.

 Public classSolution { Public BooleanSearchmatrix (int[] Matrix,inttarget) {        intLen1 =matrix.length; if(Len1 = = 0 )            return false; intLen2 = matrix[0].length; if(Len2 = = 0 | | Target < MATRIX[0][0])            return false; intRow_start = 0,row_end = Len1-1; intCol_start = 0,col_end = len2-1; introw = (Row_end+row_start)/2, col;  while(Row_start <=row_end) {Row= (Row_end+row_start)/2; if(Target >= matrix[row][0] ){                if(row = = Len1-1 | | Target < MATRIX[ROW+1][0])                     Break; ElseRow_start= Row+1; }            Else{                if(row = = 0 | | target >= matrix[row-1][0]) {row--;  Break; }                ElseRow_end= Row-1; }        }         while(Col_start <=col_end) {Col= (Col_end+col_start)/2; if(Target >Matrix[row][col]) {                if(col = =col_end)return false; ElseCol_start= Col+1; }Else if(Target <Matrix[row][col]) {                if(col = =Col_start)return false; ElseCol_end= Col-1; }Else                return true; }        return false; }}

 Public classSolution { Public BooleanSearchmatrix (int[] Matrix,inttarget) {        intLen1 =matrix.length; if(Len1 = = 0 )            return false; intLen2 = matrix[0].length; if(Len2 = = 0 | | Target < MATRIX[0][0])            return false; intStart = 0,end = len1*len2-1, Flag;  while(Start <=end) {Flag= (start+end)/2; intnum = matrix[flag/len2][flag%Len2]; if(Target >num) Start= Flag+1; Else if(Target <num) End= Flag-1; Else                return true; }        return false; }}

Leetcode Search a 2D Matrix-----java