Dungeon Game

题目连接： https://leetcode.com/problems/dungeon-game/description/

题目分析： 使用一个数组health[][]记录Knight在每一个方格所需要 的最小health值，使用动态规划，通过动态转换方程：health[i][j] = min(health[i][j + 1], health[i + 1][j]) - dungeon[i][j] <= 0 ? 1 : min(health[i][j + 1], health[i + 1][j]) - dungeon[i][j]，从dungeon的右下向左上进行状态的迁移，最终，Knight所需要的最小health值为health[0][0]。

代码：

class Solution {public:   int calculateMinimumHP(vector<vector<int> >& dungeon) {            int m = dungeon.size(), n = dungeon[0].size();            int health[m][n];            health[m-1][n-1] = dungeon[m-1][n-1] > 0 ? 1 : abs(dungeon[m-1][n-1]) + 1;            for (int i = 0; i < n-1; ++i) {                int need = health[m-1][i+1] - dungeon[m-1][i];                health[m-1][i] = need <= 0 ? 1 : abs(need);            }            for (int i = 0; i < m-1; ++i) {                int need = health[i+1][n-1] - dungeon[i][n-1];                health[i][n-1] = need <= 0 ? 1 : abs(need);            }             for (int i = m-2; i >= 0; --i)                 for (int j = n-2; j >= 0; --j) {                    int need = min(health[i][j+1], health[i+1][j]) - dungeon[i][j];                    health[i][j] = need <= 0 ? 1 : abs(need);                }            for (int i = 0; i < m; ++i)                 for (int j = 0; j < n; ++j)                    cout << health[i][j] << " ";           return health[0][0];       }};

复杂度： O(m*n).