Cost Matrix - Zenefits


http://www.1point3acres.com/bbs/thread-145290-1-1.html
/*一个正方形的矩阵,左上角的位置保证是0,其他的每个坐标位置上都有一个大于等于0的数。
* 一个人最初携带着的钱数为K,然后要从左上角走到右下角,每次只能向右或者向下走,
* 每到一个位置上就需要花掉该位置上对应的钱数,要求编程看看这个人能不能走到右下角,
* 如果不能,就返回-1,如果能,那就要找到一个解法,使得这个人剩余的钱数尽可能少,
* 也就是最终剩余的钱数要大于等于0,并且尽可能接近于0.*/

  1.         public int findMaxSpent(int[][] mat, int k){
  2.                 // dp[i][j][n] is the max mount spent at i,j but less than n. 1point 3acres 璁哄潧
  3.                 // dp[i][j][n] = max(dp[i - 1][j][n - mat[i][j]], dp[i][j - 1][n - mat[i][j]])
  4.                 int m = mat.length;. more info on 1point3acres.com
  5.                 if (m == 0) return -1;
  6.                 int n = mat[0].length; 鏉ユ簮涓€浜�.涓夊垎鍦拌鍧�. 
  7.                 int[][] dp1 = new int[n + 1][k + 1];
  8.                 int[][] dp2 = new int[n + 1][k + 1];
  9.                 // when i == -1; edge case
  10.                 for (int j = 0; j < n; j++){
  11.                         for (int h = 0; h <= k; h++){
  12.                                 dp1[j + 1][h] = -1;-google 1point3acres
  13.                         }
  14.                 }
  15.                 // when j = -1 edge case
  16.                 for (int h = 0; h <= k; h++){
  17.                         dp2[0][h] = -1;
  18.                 }
  19.                 .鐣欏璁哄潧-涓€浜�-涓夊垎鍦�
  20.                 for (int i = 0; i < m; i++){
  21.                         for (int j = 0; j < n; j++){
  22.                                 for (int h = 0; h <= k; h++){
  23.                                         if (i == 0 && j == 0) dp2[j + 1][h] = 0;
  24.                                         else if (h < mat[i][j]) dp2[j + 1][h] = -1;. From 1point 3acres bbs
  25.                                         else {
  26.                                                 int pre1 = dp2[j][h - mat[i][j]];. 鐗涗汉浜戦泦,涓€浜╀笁鍒嗗湴
  27.                                                 int pre2 = dp1[j + 1][h - mat[i][j]];
  28.                                                 if (pre1 == -1 && pre2 == -1) dp2[j + 1][h] = -1;
  29.                                                 else dp2[j + 1][h] = Math.max(pre1, pre2) + mat[i][j];
  30.                                         }
  31.                                 }
  32.                         }. Waral 鍗氬鏈夋洿澶氭枃绔�,
  33.                         dp1 = Arrays.copyOf(dp2, n + 1);
  34.                 }. 鍥磋鎴戜滑@1point 3 acres
  35.                 return dp1[n][k];
  36.         }
  37.         . 鍥磋鎴戜滑@1point 3 acres
  38.         public static void main(String[] args){
  39.                 Solution sln = new Solution();
  40.                 int[][] mat = {{0,4,5}, {1,3,2}, {0,1,1}};. from: 1point3acres.com/bbs 
  41.                 System.out.println(sln.findMaxSpent(mat, 12));
  42.         }
  43. }

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