Given an integer matrix, find the length of the longest increasing path.
From each cell, you can either move to four directions: left, right, up or down. You may NOT move diagonally or move outside of the boundary (i.e. wrap-around is not allowed).
Example 1:
Input: nums =
[
[9,9,4],
[6,6,8],
[2,1,1]
]
Output: 4
Explanation: The longest increasing path is[1, 2, 6, 9].
Example 2:
Input: nums =
[
[3,4,5],
[3,2,6],
[2,2,1]
]
Output: 4
Explanation: The longest increasing path is [3, 4, 5, 6]. Moving diagonally is not allowed.
思路就是dfs搜索加动态存储
但是对数据的处理及存储方式不同会对运行速度造成很大影响。下面列出两种方法，第一种是目前details区最快速的算法，第二种是自己写的。在本机的运行速度分别为10ms与50ms.
两种方法对于数据处理的区别主要在（用fast代表快速算法，slow同理）：
fast创建包含一位数字的数组用来存储最终结果（用数组是为了保证值的传递性），并在深度搜索的过程中随时更新值
而slow在最后再次遍历dp数组找最大值；
fast在向四个方向扩展过程中不断更新当前最大方向所对应的最大深度
而slow写了一个函数取四个方向中的最大深度；
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solution 1：
public int longestIncreasingPath(int[][] matrix) {
if(matrix == null || matrix.length == 0 || matrix[0].length == 0) {
return 0;
}
int rows = matrix.length;
int cols = matrix[0].length;
int[] global = new int[]{1};
int[][] dp = new int[rows][cols];
for(int i = 0; i < rows; i++) {
for(int j = 0; j < cols; j++) {
dfs(matrix, i, j, dp, global);
}
}
return global[0];
}
private int dfs(int[][] matrix, int row, int col, int[][] dp, int[] global) {
if(dp[row][col] != 0) {
return dp[row][col];
}
int rows = matrix.length;
int cols = matrix[0].length;
int curVal = matrix[row][col];
int leftVal = col - 1 >= 0 ? matrix[row][col - 1] : Integer.MIN_VALUE;
int rightVal = col + 1 < cols ? matrix[row][col + 1] : Integer.MIN_VALUE;
int topVal = row - 1 >= 0 ? matrix[row - 1][col] : Integer.MIN_VALUE;
int bottomVal = row + 1 < rows ? matrix[row + 1][col] : Integer.MIN_VALUE;
if(curVal >= leftVal && curVal >= rightVal && curVal >= topVal && curVal >= bottomVal) {
dp[row][col] = 1;
return 1;
}
if(col - 1 >= 0 && curVal < leftVal) {
dp[row][col] = dfs(matrix, row, col - 1, dp, global) + 1;
}
if(col + 1 < cols && curVal < rightVal) {
dp[row][col] = Math.max(dp[row][col], dfs(matrix, row, col + 1, dp, global) + 1);
}
if(row - 1 >= 0 && curVal < topVal) {
dp[row][col] = Math.max(dp[row][col], dfs(matrix, row - 1, col, dp, global) + 1);
}
if(row + 1 < rows && curVal < bottomVal) {
dp[row][col] = Math.max(dp[row][col], dfs(matrix, row + 1, col, dp, global) + 1);
}
global[0] = Math.max(global[0], dp[row][col]);
return dp[row][col];
}solution 2：
class Solution {
public int longestIncreasingPath(int[][] matrix) {
if(matrix==null || matrix.length==0 || matrix[0].length==0) return 0;
int longestPath= 0, bottombound= matrix.length, rightbound= matrix[0].length;
int[][] pathMatrix= new int[matrix.length][matrix[0].length];
dfsPath(pathMatrix,matrix,bottombound,rightbound);
for(int i=0; i<bottombound; i++){
for(int j=0; j<rightbound; j++){
longestPath= longestPath>pathMatrix[i][j]? longestPath: pathMatrix[i][j];
}
}
return longestPath;
}
public void dfsPath(int[][] pathMatrix,int[][] matrix,int rowm,int colm){
for(int row=0; row<rowm; row++){
for(int col=0; col<colm; col++){
dfsPathCore(pathMatrix,matrix,rowm,colm,row,col);
}
}
}
public int dfsPathCore(int[][] pathMatrix,int[][] matrix,int rowm,int colm,int row,int col){
if(pathMatrix[row][col]!=0) return pathMatrix[row][col];
int up=0,right=0,low=0,left=0;
if(row-1>-1){
if(matrix[row][col]>matrix[row-1][col])
up= dfsPathCore(pathMatrix,matrix,rowm,colm,row-1,col);
}
if(col+1<colm){
if(matrix[row][col]>matrix[row][col+1])
right= dfsPathCore(pathMatrix,matrix,rowm,colm,row,col+1);
}
if(row+1<rowm){
if(matrix[row][col]>matrix[row+1][col])
low= dfsPathCore(pathMatrix,matrix,rowm,colm,row+1,col);
}
if(col-1>-1){
if(matrix[row][col]>matrix[row][col-1])
left= dfsPathCore(pathMatrix,matrix,rowm,colm,row,col-1);
}
pathMatrix[row][col]= 1+maxInFour(up,right,low,left);
return pathMatrix[row][col];
}
public int maxInFour(int u,int r,int b,int l){
int res= u;
if(res<r)
res=r;
if(res<b)
res=b;
if(res<l)
res=l;
return res;
}
}将slow按照fast改进：11ms
class Solution {
public int longestIncreasingPath(int[][] matrix) {
if(matrix==null || matrix.length==0 || matrix[0].length==0) return 0;
int longestPath= 0, bottombound= matrix.length, rightbound= matrix[0].length;
int[][] pathMatrix= new int[matrix.length][matrix[0].length];
int[] global= new int[]{0};
dfsPath(pathMatrix,matrix,bottombound,rightbound,global);
return global[0];
}
public void dfsPath(int[][] pathMatrix,int[][] matrix,int rowm,int colm,int[] global){
for(int row=0; row<rowm; row++){
for(int col=0; col<colm; col++){
dfsPathCore(pathMatrix,matrix,rowm,colm,row,col,global);
}
}
}
public int dfsPathCore(int[][] pathMatrix,int[][] matrix,int rowm,int colm,int row,int col,int[] global){
if(pathMatrix[row][col]!=0) return pathMatrix[row][col];
int up=0,right=0,low=0,left=0,max=0;
if(row-1>-1){
if(matrix[row][col]>matrix[row-1][col]){
up= dfsPathCore(pathMatrix,matrix,rowm,colm,row-1,col,global);
max= max>=up? max:up;
}
}
if(col+1<colm){
if(matrix[row][col]>matrix[row][col+1]){
right= dfsPathCore(pathMatrix,matrix,rowm,colm,row,col+1,global);
max= max>=right? max:right;
}
}
if(row+1<rowm){
if(matrix[row][col]>matrix[row+1][col]){
low= dfsPathCore(pathMatrix,matrix,rowm,colm,row+1,col,global);
max= max>=low? max:low;
}
}
if(col-1>-1){
if(matrix[row][col]>matrix[row][col-1]){
left= dfsPathCore(pathMatrix,matrix,rowm,colm,row,col-1,global);
max= max>=left? max:left;
}
}
pathMatrix[row][col]= 1+max;
global[0]= global[0]>=pathMatrix[row][col]? global[0]:pathMatrix[row][col];
return pathMatrix[row][col];
}
}