leetcode 488. Zuma Game(祖玛游戏)
2017-08-28 15:55
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Think about Zuma Game. You have a row of balls on the table, colored red(R), yellow(Y), blue(B), green(G), and white(W). You also have several balls
in your hand.
Each time, you may choose a ball in your hand, and insert it into the row (including the leftmost place and rightmost place). Then, if there is a group of 3 or more balls in the same color touching, remove these balls. Keep doing this until no more balls can
be removed.
Find the minimal balls you have to insert to remove all the balls on the table. If you cannot remove all the balls, output -1.
Examples:
Input: "WRRBBW", "RB"
Output: -1
Explanation: WRRBBW -> WRR[R]BBW -> WBBW -> WBB[B]W -> WW
Input: "WWRRBBWW", "WRBRW"
Output: 2
Explanation: WWRRBBWW -> WWRR[R]BBWW -> WWBBWW -> WWBB[B]WW -> WWWW -> empty
Input:"G", "GGGGG"
Output: 2
Explanation: G -> G[G] -> GG[G] -> empty
Input: "RBYYBBRRB", "YRBGB"
Output: 3
Explanation: RBYYBBRRB -> RBYY[Y]BBRRB -> RBBBRRB -> RRRB -> B -> B[B] -> BB[B] -> empty
Note:
You may assume that the initial row of balls on the table won’t have any 3 or more consecutive balls with the same color.
The number of balls on the table won't exceed 20, and the string represents these balls is called "board" in the input.
The number of balls in your hand won't exceed 5, and the string represents these balls is called "hand" in the input.
Both input strings will be non-empty and only contain characters 'R','Y','B','G','W'.
又是一道游戏题,这次是祖玛游戏。其实这道题本身是出的有问题的。如下面这个测试用例。详细见https://discuss.leetcode.com/topic/76832/standard-test-program-is-wrong/9
Input:
However, I thought the answer might be
因此,我觉得题干中应该加上限制:”球只能插入进相同颜色的球旁边" 。
我的思路很简单,我就每次看其中连续的一串数。
如 WRRBBW。
假设把 W 填满,那么看 hands 中有没有两个 W,如果有,则填满 W 并减去 hands 中的两个 W,继续递归看 RRBBW 。递归...
再假设把 R 填满,那么看 hands 中有没有一个 R,如果有,则填满 R 并减去 hands 中的一个 R,继续递归看 WBBW 。递归...
再假设把 B 填满,那么看 hands 中有没有一个 B,如果有,则填满 B 并减去 hands 中的一个 B,继续递归看 WRRW 。递归...
再假设把 W 填满,那么看 hands 中有没有两个 W,如果有,则填满 W 并减去 hands 中的两个 W,继续递归看 WRRBB。递归...
大神也用的这个思路,但是怎么就看上去比我的要更加简洁呢?
It's just like a DFS or a Backtracking solution. word is poor, just look the code.
in your hand.
Each time, you may choose a ball in your hand, and insert it into the row (including the leftmost place and rightmost place). Then, if there is a group of 3 or more balls in the same color touching, remove these balls. Keep doing this until no more balls can
be removed.
Find the minimal balls you have to insert to remove all the balls on the table. If you cannot remove all the balls, output -1.
Examples:
Input: "WRRBBW", "RB"
Output: -1
Explanation: WRRBBW -> WRR[R]BBW -> WBBW -> WBB[B]W -> WW
Input: "WWRRBBWW", "WRBRW"
Output: 2
Explanation: WWRRBBWW -> WWRR[R]BBWW -> WWBBWW -> WWBB[B]WW -> WWWW -> empty
Input:"G", "GGGGG"
Output: 2
Explanation: G -> G[G] -> GG[G] -> empty
Input: "RBYYBBRRB", "YRBGB"
Output: 3
Explanation: RBYYBBRRB -> RBYY[Y]BBRRB -> RBBBRRB -> RRRB -> B -> B[B] -> BB[B] -> empty
Note:
You may assume that the initial row of balls on the table won’t have any 3 or more consecutive balls with the same color.
The number of balls on the table won't exceed 20, and the string represents these balls is called "board" in the input.
The number of balls in your hand won't exceed 5, and the string represents these balls is called "hand" in the input.
Both input strings will be non-empty and only contain characters 'R','Y','B','G','W'.
又是一道游戏题,这次是祖玛游戏。其实这道题本身是出的有问题的。如下面这个测试用例。详细见https://discuss.leetcode.com/topic/76832/standard-test-program-is-wrong/9
Input:
"RRWWRRBBRR", "WB". The test program gave the expected answer
-1.
However, I thought the answer might be
2. Because:
RRWWRRBBRR -> RRWWRRBBR[W]R -> RRWWRRBB[B]RWR -> RRWWRRRWR -> RRWWWR -> RRR -> empty
因此,我觉得题干中应该加上限制:”球只能插入进相同颜色的球旁边" 。
我的思路很简单,我就每次看其中连续的一串数。
如 WRRBBW。
假设把 W 填满,那么看 hands 中有没有两个 W,如果有,则填满 W 并减去 hands 中的两个 W,继续递归看 RRBBW 。递归...
再假设把 R 填满,那么看 hands 中有没有一个 R,如果有,则填满 R 并减去 hands 中的一个 R,继续递归看 WBBW 。递归...
再假设把 B 填满,那么看 hands 中有没有一个 B,如果有,则填满 B 并减去 hands 中的一个 B,继续递归看 WRRW 。递归...
再假设把 W 填满,那么看 hands 中有没有两个 W,如果有,则填满 W 并减去 hands 中的两个 W,继续递归看 WRRBB。递归...
public int minSteps(String board,int[] hands){ board=removeDuplicate3(board); if(board.length()==0){ return 0; } char[] cs=board.toCharArray(); int min=6;//因为The number of balls in your hand won't exceed 5 int low=0; int high=1; int duplicate=1; while(low<cs.length){ if(high>=cs.length||cs[high]!=cs[low]){ int need=3-duplicate; if(hands[cs[low]-'A']>=need){ hands[cs[low]-'A']-=need; String subStr=""; if(high>=cs.length){ subStr=board.substring(0,low); } else{ subStr=board.substring(0,low)+board.substring(high); } int step=need+minSteps(subStr, hands); if(step<min){ min=step; } hands[cs[low]-'A']+=need; } low=high; high=low+1; duplicate=1; } else{ high++; duplicate++; } } return min; } public String removeDuplicate3(String board){ boolean isFind=true; String result=""; while(isFind==true){ isFind=false; if(board.length()<3){ return board; } result=""; int low=0; int fast=1; int duplicate=1; //之所以不让fast<board.length() //是考虑到BYYBBRRB这个末尾有一个独立的字母的情况 while(low<board.length()){ while(fast<board.length()&&board.charAt(fast)==board.charAt(low)){ fast++; duplicate++; } if(duplicate<3){ for(int i=low;i<fast;i++){ result+=board.charAt(i); } } else{ isFind=true; } low=fast; fast=low+1; duplicate=1; } board=result; } return result; }需要注意的是 removeDuplicate3 方法中需要循环,是因为存在输入的 board 是 RRGGWYYYWWGGR 的情况。每次循环去除 一串 连续超过3次的字母,接下来继续循环。
大神也用的这个思路,但是怎么就看上去比我的要更加简洁呢?
It's just like a DFS or a Backtracking solution. word is poor, just look the code.
int MAXCOUNT = 6; // the max balls you need will not exceed 6 since "The number of balls in your hand won't exceed 5" public int findMinStep(String board, String hand) { int[] handCount = new int[26]; for (int i = 0; i < hand.length(); ++i) ++handCount[hand.charAt(i) - 'A']; int rs = helper(board + "#", handCount); // append a "#" to avoid special process while j==board.length, make the code shorter. return rs == MAXCOUNT ? -1 : rs; } private int helper(String s, int[] h) { s = removeConsecutive(s); if (s.equals("#")) return 0; int rs = MAXCOUNT, need = 0; for (int i = 0, j = 0 ; j < s.length(); ++j) { if (s.charAt(j) == s.charAt(i)) continue; need = 3 - (j - i); //balls need to remove current consecutive balls. if (h[s.charAt(i) - 'A'] >= need) { h[s.charAt(i) - 'A'] -= need; rs = Math.min(rs, need + helper(s.substring(0, i) + s.substring(j), h)); h[s.charAt(i) - 'A'] += need; } i = j; } return rs; } //remove consecutive balls longer than 3 private String removeConsecutive(String board) { for (int i = 0, j = 0; j < board.length(); ++j) { if (board.charAt(j) == board.charAt(i)) continue; if (j - i >= 3) return removeConsecutive(board.substring(0, i) + board.substring(j)); else i = j; } return board; }
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