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：

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;
}
};