[leetcode-174]Dungeon Game(java)

问题描述：问题描述

分析：这道题第一眼看上去就是二维DP问题，但是我之前是从(0,0)到(row-1,col-1)的遍历过程，但是这样一来的话，就会在某一项出现问题，详见代码2；出现问题之后，我想那肯定要用DFS了，发现果然超时TLE。最后看了网友的答案是从后向前搜索，这样就很巧妙了避免了代码2中的问题。非常神奇。
递推关系式为：dp[i][j] = max(min(dp[i+1][j],dp[i][j])-dungeon[i][j]],1)

代码如下：336ms

public class Solution {       public int calculateMinimumHP(int[][] dungeon) {        int row = dungeon.length;        int col = dungeon[0].length;        dungeon[row-1][col-1] = Math.max(1-dungeon[row-1][col-1],1);        //init last row        for(int i = col-2;i>=0;i--){            dungeon[row-1][i] = Math.max(dungeon[row-1][i+1]-dungeon[row-1][i],1);        }        //init last col        for(int i = row-2;i>=0;i--){            dungeon[i][col-1] = Math.max(dungeon[i+1][col-1]-dungeon[i][col-1],1);        }        for(int i = row-2;i>=0;i--){            for(int j = col-2;j>=0;j--){                dungeon[i][j] = Math.max(Math.min(dungeon[i+1][j],dungeon[i][j+1])-dungeon[i][j],1);            }        }        return dungeon[0][0];    }}

代码2如下：

//当出现[[1,-3,3],[0,-2,0],[-3,-3,-3]]测试用例时出错//原因是这个测试用例将会导致某节点的上节点为Node(3,4)，而左节点为Node(2,1)//这样一来的话，我们不能确定当前节点应取上节点还是取下节点。因为这关系到以后的节点的选取。public class Solution {      class Node{        int maxPoint;        int curPoint;        Node(int maxPoint,int curPoint){            this.maxPoint = maxPoint;            this.curPoint = curPoint;        }        Node(Node p){            this.maxPoint = p.maxPoint;            this.curPoint = p.curPoint;        }    }    public int calculateMinimumHP(int[][] dungeon) {        //使用二维DP算法，dp[i][j] = max(dp[i-1][j],dp[i][j-1])+dungeon[i][j];        int row = dungeon.length;        int col = dungeon[0].length;        Node[][] dp = new Node[row][col];        dp[0][0] = new Node(1,1);        if(dungeon[0][0]>0){            dp[0][0].curPoint = 1+dungeon[0][0];        }else{            dp[0][0].maxPoint = 1-dungeon[0][0];        }        //init first row        for(int j = 1;j<col;j++){            dp[0][j] = new Node(dp[0][j-1]);            updatePoint(dp[0][j],dungeon[0][j]);        }        //init first col        for(int i = 1;i<row;i++){            dp[i][0] = new Node(dp[i-1][0]);            updatePoint(dp[i][0],dungeon[i][0]);        }        for(int i = 1;i<row;i++){            for(int j = 1;j<col;j++){                Node up = new Node(dp[i-1][j]);                updatePoint(up,dungeon[i][j]);                Node left = new Node(dp[i][j-1]);                updatePoint(left,dungeon[i][j]);                dp[i][j] = up.maxPoint<left.maxPoint?up:left;            }        }        return dp[row-1][col-1].maxPoint;    }    public void updatePoint(Node p,int cur){        int val = p.curPoint+cur;        if(val<1){            p.curPoint = 1;            p.maxPoint += (1-val);        }else            p.curPoint = val;    }}

代码3：小数据集可以通过，大数据集TLE

    int maxPoint=Integer.MAX_VALUE;    int[][] dungeon;    int endRow,endCol;    public int calculateMinimumHP(int[][] dungeon) {        this.dungeon = dungeon;        endRow = dungeon.length;        endCol = dungeon[0].length;        dfs(0,0,1,1);        return maxPoint;    }    private void dfs(int row,int col,int maxhp,int curhp){        int val = curhp+dungeon[row][col];        if(val<0){            maxhp+=(1-val);            curhp = 1;        }else{            curhp = val;        }        if(row+1<endRow)            dfs(row+1,col,maxhp, curhp);        if(col+1<endCol)            dfs(row,col+1,maxhp,curhp);        if(row==endRow-1&&col==endCol-1){            if(maxhp<maxPoint)                maxPoint = maxhp;        }    }