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54. Search a 2D Matrix && Climbing Stairs （Easy）
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）
// Fibonacciclass 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]; }};
54. Search a 2D Matrix && Climbing Stairs （Easy）
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