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：
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：
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;
}
};