題目描述：

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:

nums = [
  [9,9,4],
  [6,6,8],
  [2
 
,1,1]
]

Return4
The longest increasing path is[1, 2, 6, 9].

Example 2:

nums = [
  [3,4,5],
  [3,2,6],
  [2,2,1]
]

Return4
The longest increasing path is[3, 4, 5, 6]. Moving diagonally is not allowed.

對於矩陣中的每一個數字，用深度優先遍歷尋找最比當前數字大的數字，對於已經走過的數字，將其置位為INT_MIN，下次經過這個數字可以直接不用考慮。如果走到某個位置，其上下左右的數字都比當前數字小，那麼說明已經無路可走，那麼就得進行回溯，回溯的過程中必須將之前置位為INT_MIN的數字恢復為初始數字。

程式碼如下：

class Solution {
private:
    int curLen;
    int maxLen;
    int row;
    int col;
public:
    void Move(vector<vector<int> >& matrix, int x, int y)
    {
        int cur = matrix[x][y];
        matrix[x][y] = INT_MIN;
        curLen++;
        if (curLen > maxLen)
        {
            if (curLen == 8)
            {
                curLen = curLen;
            }
            maxLen = curLen;
        }
        if (y + 1 < col && matrix[x][y+1] > cur)
        {
            Move(matrix, x, y+1);
        }
        if (y - 1 >= 0 && matrix[x][y-1] > cur)
        {
            Move(matrix, x, y-1);
        }
        if (x + 1 < row && matrix[x+1][y] > cur)
        {
            Move(matrix, x+1, y);
        }
        if (x - 1 >= 0 && matrix[x-1][y] > cur)
        {
            Move(matrix, x-1, y);
        }
        matrix[x][y] = cur;
        curLen--;
    }
    int longestIncreasingPath(vector<vector<int> >& matrix) {
        if (matrix.empty()||matrix[0].empty()) return 0;
        row = matrix.size();
        col = matrix[0].size();

        for (int i = 0; i < row; i++)
        {
            for (int j = 0; j < col; j++)
            {
                curLen = 0;
                Move(matrix, i, j);
            }
        }

        return maxLen;
    }
};