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题目描述:
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:给定 m x n 二维整型数组,一个目标值target,数组满足两个性质:① 元素每行递增;② 每行首元素大于前一行末元素。判断目标值在数组中是否存在。
思路:根据二维整型数组的两个有序特性,可以采用二分法。① 先考察最后一列元素,对于列matrix[0][n - 1]→matrix[m - 1][n - 1]进行二分查找(指针初始化left = 0,right = m - 1),若找到,则返回true;否则在指针left指向的行内查找(根据性质②)。② 对于行matrix[left][0]→matrix[left][n - 1]进行二分查找(指针初始化l = 0,r = n - 1),若找到,则返回true;否则返回false。
时间复杂度为O(log(m)) + O(log(n)) = O(log(m * n))。
代码:
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
// Exceptional Case:
if(matrix.empty()){
return false;
}
int m = matrix.size(), n = matrix[0].size();
// for case [[]]: m = 1, n = 0
if(m == 0 || n == 0){
return false;
}
// debug
// cout << "m: " << m << ", n: " << n << endl;
if(target < matrix[0][0] || target > matrix[m - 1][n - 1]){
return false;
}
// binary search
int left = 0, right = m - 1;
while(left <= right){
int mid = (left + right) >> 1;
if(matrix[mid][n - 1] == target){
return true;
}
else if(matrix[mid][n - 1] > target){
right = mid - 1;
}
else if(matrix[mid][n - 1] < target){
left = mid + 1;
}
}
int l = 0, r = n - 1;
while(l <= r){
int m = (l + r) >> 1;
if(matrix[left][m] == target){
return true;
}
else if(matrix[left][m] > target){
r = m - 1;
}
else if(matrix[left][m] < target){
l = m + 1;
}
}
return false;
}
};