leetcode:Search a 2D Matrix（数组，二分查找)

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.

分析：题意为在一个mxn矩阵中查找目标值。可以先通过二分法确定目标值target可能出现的行，然后再用一次二分法确定目标值target在行中的可能位置。

 时间复杂度为O(logn+logm)

code如下：

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class Solution { public:     bool searchMatrix(vector<vector<int>>& matrix, int target) {         int left=0;         int right=matrix.size()-1;         if(left != right){             while(left <= right){                 int mid=left + (right-left)/2;                 if(matrix[mid][0]<target){                     left=mid+1;                 }                 else if(matrix[mid][0]>target){                     right=mid-1;                 }                 else {                     return true;                 }             }         }         if(right==-1){             return false;         }         else{             int row=right;             int left=0;             int right=matrix[row].size()-1;             while(left<=right){                 int mid=left + (right-left)/2;                 if(matrix[row][mid]<target){                     left=mid+1;                 }                 else if(matrix[row][mid]>target){                     right=mid-1;                 }                 else {                     return true;                 }             }             return false;         }     } };

其他思路：

从左下角元素开始遍历，每次遍历中若与目标值target相等则返回true；若小于则列向右移动；若大于则行向下移动。时间复杂度O(logn+logm)

code如下：

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class Solution { public:     bool searchMatrix(vector<vector<int>>& matrix, int target) {         int i=matrix.size()-1;         int j=0;         int m=matrix.size();         int n=matrix[0].size();         while(i>=0 && j<n){             if(matrix[i][j] > target){                 i--;             }             else if(matrix[i][j] == target){                 return true;             }             else{                 j++;             }         }         return false;     }      };