leetcode动态规划笔记二---矩阵型DP
动态规划
题目分类
矩阵型DP
- Unique Paths II : 矩阵型DP，求所有方法总数
- Minimum Path Sum:矩阵型，求最大最小值
- Triangle : 矩阵型，求最大最小值
- Maximum Square ：矩阵型，求最大最小值
- Range Sum Query 2D - Immutable : 求和
Unique Paths II : 矩阵型DP，求所有方法总数
方法一：自顶向下
递归法,time limited
class Solution {
int helper(int[][] g, int x, int y){
int top = 0, left = 0;
if(g[x][y] == 1) return 0;
if(x == 0 && y == 0) return 1;
if(x > 0){
top = helper(g, x - 1, y);
}
if(y > 0){
left = helper(g, x, y - 1);
}
return top + left;
}
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
return helper(obstacleGrid, obstacleGrid.length - 1, obstacleGrid[0].length -1);
}
}
方法一：自顶向下
递归法 + memo ac
class Solution {
Map<Long, Integer> m = new HashMap<>();
int helper(int[][] g, int x, int y){
int top = 0, left = 0;
if(g[x][y] == 1) return 0;
if(x == 0 && y == 0) return 1;
long key = ((long)x) << 32 | y; // long key = x << 32 | y; 错误
if(m.containsKey(key)) return m.get(key);
if(x > 0){
top = helper(g, x - 1, y);
}
if(y > 0){
left = helper(g, x, y - 1);
}
m.put(key, top + left);
return top + left;
}
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
return helper(obstacleGrid, obstacleGrid.length - 1, obstacleGrid[0].length -1);
}
}
方法三：自底向上
递推 + memo
这里用了二维memo，其实状态转移中，当前step只与上一step有关系，只用一维memo就可以了。
class Solution {
public int uniquePathsWithObstacles(int[][] g) {
if(g.length == 0) return 0;
int[][] ps = new int[g.length][g[0].length];
boolean obstacle = false;
for(int i = 0; i < g.length; i ++){
if(g[i][0] == 1){
obstacle = true;
}
ps[i][0] = obstacle ? 0 : 1;
}
obstacle = false;
for(int i = 0; i < g[0].length; i ++){
if(g[0][i] == 1){
obstacle = true;
}
ps[0][i] = obstacle ? 0 : 1;
}
for(int i = 1; i < g.length; i++){
for(int j = 1; j < g[0].length; j++){
if(g[i][j] == 1){
ps[i][j] = 0;
}
else{
ps[i][j] = ps[i-1][j] + ps[i][j-1];
}
}
}
return ps[g.length-1][g[0].length-1];
}
}
Minimum Path Sum:矩阵型，求最大最小值
方法一：自底向上 + 二维memo
class Solution {
public int minPathSum(int[][] g) {
if (g.length == 0) return 0;
int[][] sum = new int[g.length][g[0].length];
sum[0][0] = g[0][0];
for(int i = 1; i < g.length; i++){
sum[i][0] = g[i][0] + sum[i-1][0];
}
for(int i = 1; i < g[0].length; i++){
sum[0][i] = g[0][i] + sum[0][i-1];
}
for(int i = 1; i < g.length; i ++){
for(int j = 1; j < g[0].length; j++){
sum[i][j] = g[i][j] + Math.min(sum[i-1][j], sum[i][j-1]);
}
}
return sum[g.length-1][g[0].length-1];
}
}
方法二：自底向上 + 一维memo
class Solution {
public int minPathSum(int[][] g) {
if (g.length == 0) return 0;
int[] sum = new int[g[0].length];
sum[0] = g[0][0];
for(int i = 1; i < g[0].length; i++){
sum[i] = g[0][i] + sum[i-1];
}
for(int i = 1; i < g.length; i ++){
sum[0] = sum[0] + g[i][0];
for(int j = 1; j < g[0].length; j++){
sum[j] = g[i][j] + Math.min(sum[j], sum[j-1]);
}
}
return sum[g[0].length-1];
}
}
Triangle : 矩阵型，求最小值
方法一 ： 自定向下
递归 + memo
问题拆分、状态、状态转移方程、初始值，一个都不能少
class Solution {
Map<Long, Integer> m = new HashMap<>();
int helper(List<List<Integer>> tri, int row, int col){
long key = (long)row << 32 | col;
if(m.containsKey(key)){
return m.get(key);
}
if ( row == 0 && col == 0 ){ //不能忽略此逻辑
return tri.get(0).get(0);
}
int left = Integer.MAX_VALUE;
int top = Integer.MAX_VALUE;
if(row > 0){
if(col < tri.get(row).size() - 1){
top = helper(tri, row - 1, col);
}
if(col > 0){
left = helper(tri, row - 1, col - 1);
}
}
int val = Math.min(top, left) + tri.get(row).get(col);
m.put(key, val);
return val;
}
public int minimumTotal(List<List<Integer>> tri) {
if(tri.size() == 0) return 0;
int sum = Integer.MAX_VALUE;
for(int j = 0; j < tri.get(tri.size()-1).size(); j++){
sum = Math.min(sum, helper(tri, tri.size()-1, j));
}
return sum;
}
}
方法二 ： 自底向上
递推 + memo，下面的方法还可以进一步优化
class Solution {
public int minimumTotal(List<List<Integer>> tri) {
if(tri.size() == 0) return 0;
int[] memo = new int[tri.get(tri.size()-1).size()];
memo[0] = tri.get(0).get(0);
for(int i = 1; i < tri.size(); i ++){
for(int j = tri.get(i).size() -1; j >= 0 ; j--){
if(j == 0){
memo[j] = memo[j] + tri.get(i).get(j);
}
else if(j == tri.get(i).size() - 1){
memo[j] = memo[j-1] + tri.get(i).get(j);
}
else{
memo[j] = Math.min(memo[j-1], memo[j]) + tri.get(i).get(j);
}
}
}
int sum = Integer.MAX_VALUE;
for(int i=0; i<memo.length; i++){
sum = Math.min(sum, memo[i]);
}
return sum;
}
}
Maximum Square ：矩阵型，求最大最小值
方法一：自底向上
递推 + memo
主要看斜对角线，当前位置能成为多大square，要看左上角位置有多大square。本地弯弯道道挺多。
class Solution {
public int maximalSquare(char[][] m) {
if(m.length == 0) return 0;
int[][] memo = new int[m.length + 1][m[0].length + 1];
for(int i = 0; i <= m.length; i++ ){
memo[i][0] = 0;
}
for(int i = 0; i <= m[0].length; i++){
memo[0][i] = 0;
}
int maxi = 0;
for(int i = 0; i < m.length; i++){
for(int j = 0; j < m[0].length; j++){
if(m[i][j] != '1'){
memo[i+1][j+1] = 0;
continue;
}
if(memo[i][j] <= 0){
memo[i+1][j+1] = 1;
}
else {
int k = 1;
for(; k <= memo[i][j]; k++){
if(m[i-k][j] != '1' || m[i][j-k] != '1'){
break;
}
}
memo[i+1][j+1] = k;
}
//System.out.println("i" + i + " j" + j + " =" + memo[i+1][j+1]);
maxi = Math.max(memo[i+1][j+1], maxi);
}
}
return maxi * maxi;
}
}
Range Sum Query 2D - Immutable
方法一 ： 自底向上， 递推 + memo
class NumMatrix {
private int[][] memo;
public NumMatrix(int[][] m) {
if(m.length == 0 || m[0].length == 0) return;
memo = new int[m.length][m[0].length + 1];
for(int i = 0; i < m.length; i++){
for(int j = 0; j < m[0].length; j++){
memo[i][j+1] = memo[i][j] + m[i][j];
}
}
}
public int sumRegion(int row1, int col1, int row2, int col2) {
int sum = 0;
for(int i = row1; i <= row2; i++){
sum += (memo[i][col2+1] - memo[i][col1]);
}
return sum;
}
}
