标签：style blog color io ar strong for art div
Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
For example,
Consider the following matrix:
[ [1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50] ]
Given target = 3, return true.
写得蛮啰嗦，简而言之就是先搜索列号，再在搜索到的列里面去搜索该元素，全部用的二分查找（反正它已经排序了）
1 public boolean searchMatrix(int[][] matrix, int target) { 2 if(matrix==null||matrix.length <=0){ 3 return false; 4 } 5 int targetRow =0; 6 int m = matrix.length; 7 int n = matrix[0].length; 8 9 //make sure target is in the range 10 if(target < matrix[0][0] || target > matrix[m-1][n-1]){ 11 return false; 12 }else{ 13 targetRow = searchRow(matrix,target,0,m-1); 14 if(targetRow == -1){ 15 return false; 16 } 17 return searchElement(matrix,target,targetRow,0,n-1); 18 } 19 20 } 21 public int searchRow(int[][] matrix,int target,int start,int end){ 22 while(start <= end){ 23 int mid = (end-start)/2 +start; 24 if(matrix[mid][0] == target){ 25 return mid; 26 } 27 if(matrix[mid][0] < target){ 28 if(matrix[mid][matrix[0].length -1] >= target){ 29 return mid; 30 }else{ 31 start = mid +1; 32 continue; 33 } 34 } 35 if(matrix[mid][0] > target){ 36 end = mid-1; 37 continue; 38 } 39 40 } 41 return -1; 42 } 43 44 public boolean searchElement(int [][] matrix,int target,int row,int start,int end){ 45 while(start <= end){ 46 int mid = (end-start)/2 + start; 47 if( matrix[row][mid] == target){ 48 return true; 49 } 50 if(matrix[row][mid] < target){ 51 start = mid+1; 52 continue; 53 } 54 if(matrix[row][mid] > target){ 55 end = mid - 1; 56 continue; 57 } 58 } 59 return false; 60 }
标签：style blog color io ar strong for art div
原文地址：http://www.cnblogs.com/dijkstra-c/p/3957660.html
