• 329. Longest Increasing Path in a Matrix


    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.
    public class Solution {
        public int longestIncreasingPath(int[][] matrix) {
            if(matrix==null||matrix.length==0) return 0;
            int row=matrix.length,col=matrix[0].length;
            int result[][]=new int[row][col]; //store the longest path at every node
            int max=1;
            for(int[]arr:result){
                Arrays.fill(arr,-1);
            }
            for(int i=0;i<row;i++){
                for(int j=0;j<col;j++){
                    findPath(matrix,result,i,j);
                    max=Math.max(result[i][j],max);
                }
            }
            return max;
        }
        
        private void findPath(int[][]matrix,int result[][],int i,int j){
            int row=matrix.length,col=matrix[0].length;
            
            int xPos[]={0,0,1,-1};
            int yPos[]={1,-1,0,0};
            int max=1;
            for(int k=0;k<4;k++){  //at four direction
                
                    int nX=xPos[k]+i;
                    int nY=yPos[k]+j;
                    if(nX>=row||nX<0||nY>=col||nY<0){ //skip when out of boundary
                        continue;
                    }
                    if(matrix[nX][nY]>matrix[i][j]){
                        if(result[nX][nY]<0){  //haven't visited [nX,nY]
                            findPath(matrix,result,nX,nY);
                        }
                        max=Math.max( 1+result[nX][nY],max);
                    }
                
            }
           result[i][j]=max;
            
        }
    }

    答案一:用数组表示当前元素拥有的最大增长path,上下左右四个方向用0,±1表示,很巧妙,然后是dfs

    class Solution {
        public int longestIncreasingPath(int[][] matrix) {
            int max = 0;
            if(matrix.length == 0)
                return 0;
            int[][] holder = new int[matrix.length][matrix[0].length];
            for(int i = 0; i < matrix.length; i++){
                for(int j = 0; j < matrix[0].length; j++){
                    int _max = pathFinder(matrix, holder, Integer.MIN_VALUE, i, j);
                    max = _max > max ? _max : max;
                }
            }
            return max;
        }
        
        
        private int pathFinder(int[][] matrix, int[][] dp, int last, int v, int w){
            if(v < 0 || w < 0 || v == matrix.length || w == matrix[0].length || matrix[v][w] <= last)
                return 0;
            if(dp[v][w] > 0){
                return dp[v][w];
            }
            int val = pathFinder(matrix, dp, matrix[v][w], v-1, w);
            val = Math.max(pathFinder(matrix, dp, matrix[v][w], v+1, w), val);
            val = Math.max(pathFinder(matrix, dp, matrix[v][w], v, w-1), val);
            val = Math.max(pathFinder(matrix, dp, matrix[v][w], v, w+1), val);
            dp[v][w] = ++val;
            return val;
        }
    }

    答案2更厉害了,但是本质和1一样

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  • 原文地址:https://www.cnblogs.com/wentiliangkaihua/p/11811268.html
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