Search a 2D Matrix -- Leetcode
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12.25 2014
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 from left to right.
- The first integer of each row is greater than the last integer of the previous row.
For example,
Consider the following matrix:
[ [1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50]]
Given target = 3
, return true
.
class Solution {public: bool searchMatrix(vector<vector<int> > &matrix, int target) { int i=0; int n=matrix.size(); int m=matrix[0].size(); int row=0; while(i<n){ if(target==matrix[i][0]) return true; else if(target>matrix[i][0]){ row=i; i++; } else if(target<matrix[i][0] && i!=0){ if(target>matrix[i-1][m-1]) return false; else if(target==matrix[i-1][m-1]) return true; else{ row=i-1; break; } } else if(target<matrix[i][0] && i==0) return false; } int begin=0; int end=m-1; while(begin<=end){ int mid=(begin+end)/2; if(matrix[row][mid]==target) return true; else if(matrix[row][mid]<target) begin=mid+1; else end=mid-1; } return false; }};
第一步是先确定row的位置。细节实现一下即可,根据不同的情况分类讨论一下,唯一要注意的是当i是零时,要特别处理,此时若target小于第一排的第一个数,那么肯定无法在该matrix里面找到。在确定了行的情况下,用二分法查找。该步实现时要注意,while循环条件有所不同,是begin<=end,因为可能begin和end会重合且可能该位置就是target的位置。后来转念一下为何不两步都使用二分法呢?
class Solution {public: bool searchMatrix(vector<vector<int> > &matrix, int target) { int i=0; int n=matrix.size(); int m=matrix[0].size(); int begin=0; int end=n-1; while(begin<=end){ int mid=(begin+end)/2; if(matrix[mid][0]==target) return true; else if(matrix[mid][0]<target) begin=mid+1; else end=mid-1; } int row=end<0?begin:end; begin=0; end=m-1; while(begin<=end){ int mid=(begin+end)/2; if(matrix[row][mid]==target) return true; else if(matrix[row][mid]<target) begin=mid+1; else end=mid-1; } return false; }};不同在于第一步。while条件仍为begin<=end。需要注意的地方时row的取值。一般条件下row的取值为end,但是考虑到第一排的关系,所以还应该判断end是否为-1;
第一个代码跑了86ms 第二个跑了56ms 时间性能提升
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