LeetCode 62

/*
 * @lc app=leetcode id=62 lang=cpp
 *
 * [62] Unique Paths
 *
 * https://leetcode.com/problems/unique-paths/description/
 *
 * algorithms
 * Medium (46.62%)
 * Total Accepted:    268.7K
 * Total Submissions: 574.4K
 * Testcase Example:  '3
2'
 *
 * A robot is located at the top-left corner of a m x n grid (marked 'Start' in
 * the diagram below).
 * 
 * The robot can only move either down or right at any point in time. The robot
 * is trying to reach the bottom-right corner of the grid (marked 'Finish' in
 * the diagram below).
 * 
 * How many possible unique paths are there?
 * 
 * 
 * Above is a 7 x 3 grid. How many possible unique paths are there?
 * 
 * Note: m and n will be at most 100.
 * 
 * Example 1:
 * 
 * 
 * Input: m = 3, n = 2
 * Output: 3
 * Explanation:
 * From the top-left corner, there are a total of 3 ways to reach the
 * bottom-right corner:
 * 1. Right -> Right -> Down
 * 2. Right -> Down -> Right
 * 3. Down -> Right -> Right
 * 
 * 
 * Example 2:
 * 
 * 
 * Input: m = 7, n = 3
 * Output: 28
 * 
 */
class Solution {
public:
    int uniquePaths(int m, int n) {
       if (m == 1 && n ==1)
            return 1; 
        int right = 0,left = 0;
        if(m > 1) right = uniquePaths(m-1,n);
        if(n > 1) left = uniquePaths(m,n-1);
        return right + left;
    }
};

class Solution {
public:
    int uniquePaths(int m, int n) {
        int ways[m][n];
        for(int i = 0;i < m;++i){
            for(int j = 0;j < n; ++j){
                if(i == 0 || j == 0) ways[i][j] = 1;
                else
                {
                    ways[i][j] = ways[i-1][j] + ways[i][j-1];
                }                
            }
        }
        return ways[m-1][n-1];
    }
};

class Solution {
public:
    int uniquePaths(int m, int n) {
        int ways[n]={0};
        ways[0] = 1;
        for(int i = 0;i < m;++i){
            for(int j = 1;j < n; ++j){
                ways[j] += ways[j-1];             
            }
        }
        return ways[n-1];
    }
};

首先是朴素的递归思想；
其次是根据动态规划的状态转移方程进行优化；
再对存储进行优化；