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POfind an element in a sorted matrix
 

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    <p>Problem: Given a matrix in which each row and each column is sorted, write a method to find an element in it. </p> <p>It is a classic interview question, here is my solution</p> <pre><code>boolean F(int[][] matrix, int hs, int he, int ws, int we) { if (hs &gt; he || ws &gt; we) return false; int m = (hs + he) / 2; int n = (ws + we) / 2; if (matrix[m][n] == t) { return true; } else if (matrix[m][n] &lt; t) { // find the ele in the same row, right to [m][n] F(m, m, n + 1, we); // find the ele in the same col, upper to [m][n] F(m + 1, he, n, n); // find the ele in the area, where i&gt;m,j&gt;n F(m + 1, he, n + 1, we); } else if (matrix[m][n] &gt; t) { // very similar to previous part } } </code></pre> <p>The running time of the algorithm is log(m) + log(n). I am looking for an algorithm that is more efficient, or with concise code.</p> <p>Having more comments, I come up with following code:</p> <pre><code>// return target recurrence in the matrix int F(int[][] m, int rs, int re, int cs, int ce, int t){ int r1 = rs, r2 = re; int c1 = cs, c2 = ce; int r=0 , c = c1; while( r1 &lt; r2 &amp;&amp; c1 &lt; c2 ){ // find the last element that &lt;= t in column c r = FlastLess( r1, r2, c, t) if( r == -1 ) break; else{ // find the first ele in the row that is &gt;=t c = FfirstGreater( r, c1, c2, t); if( c == -1) break; else{ r2 = r; c1 = c; }// else }// else }// while }// f </code></pre> <p>Here is the link to function F1 and F2 <a href="https://stackoverflow.com/questions/6553970/find-the-first-element-in-an-array-that-is-greater-than-the-target">Find the first element in a sorted array that is greater than the target</a></p> <pre><code>void FlastLess(int s, int e, int t){ int l = s, h = e; while( l != h ){ int mid = (l+h)/2; if( mid &gt;= t) high = mid - 1; else { if( high &lt; t) low= mid + 1; else low = mid; } } void FfirstGreater(int s, int e, int t){ while(l &lt; h){ mid = (l+h)/2; if ( mid &lt;= t) low = mid+1; else high = mid; } } } </code></pre>
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    1. COI could be wrong, but binary search might be as fast as you can get it.
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    2. COI'm not sure your method works. Imagine a matrix whose first row is [0,10,20,..,90] and the next row is [1,11,21,..,91] up to [9,19,29,...,99]. In this case each row and column is ordered. Now you start from 55, and you're looking for 72. 72 > 55 but it's not in the bottom half of the matrix. If you're looking for 19, it's not in the top half. Maybe I didn't understand the algorithm. I also don't understand how you have consecutive "return" statements - it's unreachable code.
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    3. CO@secureFish consider new answer . And let me know it is right .
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