leetcode 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.

Note: 

(1) You may imagine 

nums[-1] = nums
= 1

. They are not real therefore you can not burst
them.

(2) 0 ≤ 

n

 ≤ 500, 0 ≤ 

nums[i]

 ≤
100

Example:

Given 

[3, 1, 5, 8]

Return 

167

nums = [3,1,5,8] --> [3,5,8] -->   [3,8]   -->  [8]  --> []
coins =  3*1*5      +  3*5*8    +  1*3*8      + 1*8*1   = 167

原文地址（我的新blog）：http://www.hrwhisper.me/leetcode-burst-balloons/

题意：

给定n个气球。每次你可以打破一个，打破第i个，那么你会获得nums[left] * nums[i] * nums[right]个积分。 （nums[-1] = nums = 1）求你可以获得的最大积分数

思路:

看了discuss是dp[i][j]为打破的气球为i~j之间。

我们可以想象：最后的剩下一个气球为i的时候，可以获得的分数为：nums[-1]*nums[i]*nums
.

那么介于i,j之间的x，有： dp[i][j] = max(dp[i][j], dp[i][x – 1] + nums[i – 1] * nums[x] * nums[j + 1] + dp[x + 1][j]);

C++

class Solution {
public:
int maxCoins(vector<int>& nums) {
int n = nums.size();
nums.insert(nums.begin(), 1);
nums.insert(nums.end(), 1);
vector<vector<int> > dp(n + 2, vector<int>(n + 2, 0));
for (int k = 1; k <= n; k++) {
for (int i = 1; i <= n - k + 1; i++) {
int j = i + k - 1;
for (int x = i; x <= j; x++) {
int temp = dp[i][x - 1] + nums[i - 1] * nums[x] * nums[j + 1] + dp[x + 1][j];
if (dp[i][j] < temp)
dp[i][j] = temp;
}
}
}
return dp[1]
;
}
};

java

public class Solution {
public int maxCoins(int[] iNums) {
int n = iNums.length;
int[] nums = new int[n + 2];
for (int i = 0; i < n; i++) nums[i + 1] = iNums[i];
nums[0] = nums[n + 1] = 1;
int[][] dp = new int[n + 2][n + 2];
for (int k = 1; k <= n; k++) {
for (int i = 1; i <= n - k + 1; i++) {
int j = i + k - 1;
for (int x = i; x <= j; x++) {
dp[i][j] = Math.max(dp[i][j], dp[i][x - 1] + nums[i - 1] * nums[x] * nums[j + 1] + dp[x + 1][j]);
}
}
}
return dp[1]
;
}
}

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