64. Minimum Path Sum
Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.
Note: You can only move either down or right at any point in time.
class Solution {
public:
int minPathSum(vector<vector<int>>& grid) {
int m = grid.size();
int n = grid[0].size();
vector<vector<int>> f(m,vector<int>(n,0));
f[0][0] = grid[0][0];
for(int i=1;i<m;i++)
f[i][0] = f[i-1][0] + grid[i][0];
for(int j=1;j<n;j++)
f[0][j] = f[0][j-1] + grid[0][j];
for(int i=1;i<m;i++)
for(int j=1;j<n;j++)
{
f[i][j] = min(f[i-1][j],f[i][j-1]) + grid[i][j];
}
return f[m-1][n-1];
}
};
62. Unique Paths
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 3 x 7 grid. How many possible unique paths are there?
Note: m and n will be at most 100.
class Solution {
public:
int uniquePaths(int m, int n) {
vector<vector<int>> f(100,vector<int>(100,0));
f[0][0] = 1;
for(int i=1;i<m;i++)
f[i][0] = f[i-1][0];
for(int j=1;j<n;j++)
f[0][j] = f[0][j-1];
for(int i=1;i<m;i++)
for(int j=1;j<n;j++)
{
f[i][j] = f[i][j-1] + f[i-1][j];
}
return f[m-1][n-1];
}
};
63. Unique Paths II
Follow up for "Unique Paths":
Now consider if some obstacles are added to the grids. How many unique paths would there be?
An obstacle and empty space is marked as 1 and 0 respectively in the grid.
For example,
There is one obstacle in the middle of a 3x3 grid as illustrated below.
[
[0,0,0],
[0,1,0],
[0,0,0]
]
The total number of unique paths is 2.
Note: m and n will be at most 100.
class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
int m = obstacleGrid.size();
int n = obstacleGrid[0].size();
vector<vector<int>> f(m,vector<int>(n,0));
if(obstacleGrid[0][0]==0)
f[0][0] = 1;
else
f[0][0] = 0;
for(int i=1;i<m;i++)
{
if(obstacleGrid[i][0]==0)
f[i][0] = f[i-1][0];
else
f[i][0] = 0;
}
for(int j=1;j<n;j++)
{
if(obstacleGrid[0][j]==0)
f[0][j] = f[0][j-1];
else
f[0][j] = 0;
}
for(int i=1;i<m;i++)
for(int j=1;j<n;j++)
{
if(obstacleGrid[i][j]==0)
f[i][j] = f[i-1][j] + f[i][j-1];
else
f[i][j] = 0;
}
return f[m-1][n-1];
}
};
72. Edit Distance
Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)
You have the following 3 operations permitted on a word:
a) Insert a character
b) Delete a character
c) Replace a character
class Solution {
public:
int minDistance(string word1, string word2) {
int m = word1.size();
int n = word2.size();
if(m<=0)
return n;
if(n<=0)
return m;
vector<vector<int>> f(m+1,vector<int>(n+1,0));
//f[i][j]表示一個(gè)長(zhǎng)為i的字符串word1變?yōu)殚L(zhǎng)為j的字符串word2的最短距離
f[0][0] = 0;//都沒有字母
for(int i=1;i<=m;i++)
{
f[i][0] = i;
}
for(int j=1;j<=n;j++)
{
f[0][j] = j;
}
for(int i=1;i<=m;i++)
for(int j=1;j<=n;j++)
{
if(word1[i-1]==word2[j-1]) //f中的第i個(gè)字母，對(duì)應(yīng)的是word1[i-1],第j個(gè)字母，對(duì)應(yīng)的是word2[j-1]
f[i][j] = f[i-1][j-1];
else
f[i][j] = min(f[i-1][j-1],min(f[i-1][j],f[i][j-1])) + 1;
}
return f[m][n];
}
};
115. Distinct Subsequences
Given a string S and a string T, count the number of distinct subsequences of T in S.
A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ACE" is a subsequence of "ABCDE" while "AEC" is not).
Here is an example:
S = "rabbbit", T = "rabbit"
Return 3.
class Solution {
public:
int numDistinct(string s, string t) {
int m = s.size();
int n = t.size();
if(m<n)
return 0;
vector<vector<int>> f(m+1,vector<int>(n+1,0));
//f[i][j]表示一個(gè)長(zhǎng)為i的字符串word1變?yōu)殚L(zhǎng)為j的字符串word2的最短距離
f[0][0] = 1;//都沒有字母
for(int i=1;i<=m;i++)
{
f[i][0] = 1;
}
for(int j=1;j<=n;j++)
{
f[0][j] = 0;
}
for(int i=1;i<=m;i++)
for(int j=1;j<=n;j++)
{
if(s[i-1]!=t[j-1]) //f中的第i個(gè)字母，對(duì)應(yīng)的是word1[i-1],第j個(gè)字母，對(duì)應(yīng)的是word2[j-1]
f[i][j] = f[i-1][j];//刪除一個(gè)字母
else
f[i][j] = f[i-1][j-1] + f[i-1][j];//都增加一個(gè)字母，或刪除s的一個(gè)字母
}
return f[m][n];
}
};
118. Pascal's Triangle
Given numRows, generate the first numRows of Pascal's triangle.
For example, given numRows = 5,
Return
[
[1],
[1,1],
[1,2,1],
[1,3,3,1],
[1,4,6,4,1]
]
class Solution {
public:
vector<vector<int>> generate(int numRows) {
vector<int> temp;
vector<vector<int>> rec(numRows,temp);
for(int i=0;i<numRows;i++)
for(int j=0;j<i+1;j++)
rec[i].push_back(1);
for(int i=1;i<numRows;i++)
for(int j=1;j<i;j++)
rec[i][j] = rec[i-1][j-1] + rec[i-1][j];
return rec;
}
};
119. Pascal's Triangle II
Given an index k, return the kth row of the Pascal's triangle.
For example, given k = 3,
Return [1,3,3,1].
class Solution {
public:
vector<int> getRow(int rowIndex) {
vector<int> temp;
vector<vector<int>> rec(rowIndex+1,temp);
for(int i=0;i<rowIndex+1;i++)
for(int j=0;j<i+1;j++)
rec[i].push_back(1);
for(int i=1;i<rowIndex+1;i++)
for(int j=1;j<i;j++)
rec[i][j] = rec[i-1][j-1] + rec[i-1][j];
return rec[rowIndex];
}
};
120. Triangle
Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.
For example, given the following triangle
[
[2],
[3,4],
[6,5,7],
[4,1,8,3]
]
The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11).
class Solution {
public:
int minimumTotal(vector<vector<int>>& triangle) {
int m = triangle.size();
int result = INT_MAX;
vector<vector<int>> f(m,vector<int>(m,0));
f[0][0] = triangle[0][0];
for(int i=1;i<m;i++)
{
f[i][0] = f[i-1][0] + triangle[i][0];
f[i][i] = f[i-1][i-1] + triangle[i][i];
cout<<f[i][0]<<"--"<<f[i][i]<<endl;
}
for(int i=1;i<m;i++)
for(int j=1;j<i;j++) //j只能到i-1
{
f[i][j] = min(f[i-1][j-1],f[i-1][j]) + triangle[i][j];
}
for(int k=0;k<m;k++)
{
if(f[m-1][k]<result)
result = f[m-1][k];
cout<<f[m-1][k]<<" ";
}
return result;
}
};
