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  • [LeetCode 312] Burst Balloons

    Given n balloons, indexed from 0 to n-1. Each balloon is painted with a number on it represented by array nums. You are asked to burst all the balloons. If the you burst balloon i you will get nums[left] * nums[i] * nums[right] coins. Here left and right are adjacent indices of i. After the burst, the left and right then becomes adjacent.

    Find the maximum coins you can collect by bursting the balloons wisely.


    - You may imagine nums[-1] = nums[n] = 1. They are not real therefore you can not burst them.
    - 0 ≤ n ≤ 500, 0 ≤ nums[i] ≤ 100

    Example

    Given [4, 1, 5, 10]
    Return 270

    nums = [4, 1, 5, 10] burst 1, get coins 4 * 1 * 5 = 20
    nums = [4, 5, 10]    burst 5, get coins 4 * 5 * 10 = 200 
    nums = [4, 10]       burst 4, get coins 1 * 4 * 10 = 40
    nums = [10]          burst 10, get coins 1 * 10 * 1 = 10
    
    Total coins 20 + 200 + 40 + 10 = 270

    Dynamic Programming Solution

    A straightforward solution is to use recursion. Take the given example[4, 1, 5, 10].

    The following recursion tree shows that the recursive approach has an overlapping subproblems issue. 

                                  4, 1, 5, 10

          1, 5, 10            4, 5, 10                4, 1, 10,                 4, 1, 5        

    5, 10             1, 10           1, 5    5, 10         4, 10       4,5         1, 10  4,10   4,1        1,5    4,5     4,1 

    Dynamic Programming Solution

    State: T[i][j] stores the max value of bursting ballons i to j and the last ballon that is bursted to get this max.

    Function: T[i][j].maxV = max{T[i][k - 1].maxV + T[k + 1][j].maxV + arr[i - 1] * arr[k] * arr[j + 1]};

         T[i][j].lastBurstIdx is the k in [i, j] that gives the max value to T[i][j].maxV.

    Init: T[i][i].maxV = arr[i - 1] * arr[i] * arr[i + 1]; T[i][j].lastBurstIdx = i.

     1 import java.util.Stack;
     2 
     3 class Entry {
     4     int maxV;
     5     int lastBurstIdx;
     6     Entry(int v, int i) {
     7         this.maxV = v;
     8         this.lastBurstIdx = i;
     9     }
    10 }
    11 public class BurstBallon {
    12     private Stack<Integer> burstSeq;
    13     public int getMaxPoint(int[] arr) {
    14         burstSeq = new Stack<Integer>();
    15         if(arr == null || arr.length == 0) {
    16             return 0;
    17         }
    18         Entry[][] T = new Entry[arr.length][arr.length];
    19         for(int i = 0; i < T.length; i++) {
    20             for(int j = i; j < T.length; j++) {
    21                 T[i][j] = new Entry(0, -1);
    22             }
    23         }
    24         for(int i = 0; i < T.length; i++) {
    25             int left = i - 1 >= 0 ? arr[i - 1] : 1;
    26             int right = i + 1 < T.length ? arr[i + 1] : 1;
    27             T[i][i].maxV = left * arr[i] * right;
    28             T[i][i].lastBurstIdx = i;
    29         }
    30         for(int len = 2; len <= arr.length; len++) {
    31             for(int i = 0; i <= arr.length - len; i++) {
    32                 int j = i + len - 1;
    33                 for(int lastBurst = i; lastBurst <= j; lastBurst++) {
    34                     int outerLeft = (i - 1) >= 0 ? arr[i - 1] : 1;
    35                     int outerRight = (j + 1) < arr.length ? arr[j + 1] : 1;
    36                     int midValue = outerLeft * arr[lastBurst] * outerRight;
    37                     int leftMax = i <= lastBurst - 1 ? T[i][lastBurst - 1].maxV : 0;
    38                     int rightMax = lastBurst + 1 <= j ? T[lastBurst + 1][j].maxV : 0;
    39                     int currValue = leftMax + midValue + rightMax;
    40                     if(currValue > T[i][j].maxV) {
    41                         T[i][j].maxV = currValue;
    42                         T[i][j].lastBurstIdx = lastBurst;
    43                     }
    44                 }
    45             }
    46         }
    47         //reconstruct one burst sequence that gives the best result
    48         reconstructOneBestSequence(T, 0, arr.length - 1, burstSeq);
    49         return T[0][arr.length - 1].maxV;
    50     }
    51     private void reconstructOneBestSequence(Entry[][] T, int start, int end, Stack<Integer> burstSeq) {
    52         if(end < start) {
    53             return;
    54         }
    55         burstSeq.push(T[start][end].lastBurstIdx);
    56         reconstructOneBestSequence(T, start, T[start][end].lastBurstIdx - 1, burstSeq);
    57         reconstructOneBestSequence(T, T[start][end].lastBurstIdx + 1, end, burstSeq);
    58     }
    59     public static void main(String[] args) {
    60         int[] arr = {4,1,5,10};
    61         BurstBallon test = new BurstBallon();
    62         //expect 270
    63         System.out.println(test.getMaxPoint(arr));
    64         //expect 1, 2, 0, 3 
    65         while(!test.burstSeq.isEmpty()) {
    66             System.out.println(test.burstSeq.pop());
    67         }
    68     }
    69 }
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  • 原文地址:https://www.cnblogs.com/lz87/p/7288854.html
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