Search a 2D Matrix
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.
思路： 从右上方开始，若小于 target, 则往下走；若大于 target, 对改行二分查找；若等 target， 返回 true.
bool binarySearch(vector<int> &A, int target) {
int l = 0, h = A.size()-2;
while(l <= h) {
int mid = (l+h) >> 1;
if(A[mid] > target) h = mid-1;
else if(A[mid] < target) l = mid+1;
else return true;
}
return false;
}
class Solution {
public:
bool searchMatrix(vector<vector<int> > &matrix, int target) {
if(!matrix.size() || !matrix[0].size()) return false;
int row = matrix.size(), col = matrix[0].size();
for(int r = 0; r < row; ++r) {
if(matrix[r][col-1] == target) return true;
if(matrix[r][col-1] > target) return binarySearch(matrix[r], target);
}
return false;
}
};
Climbing Stairs
You are climbing a stair case. It takes n steps to reach to the top.
Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?
思路： 斐波那契。此处用动归。 还可以使用矩阵二分乘。（剑指offer: 题9）
// Fibonacci
class Solution {
public:
int climbStairs(int n) {
vector<int> f(n+1, 0);
f[0] = f[1] = 1;
for(int i = 2; i <= n; ++i) f[i] = f[i-1] + f[i-2];
return f[n];
}
};