|
|
package class46;
|
|
|
|
|
|
// 本题测试链接 : https://leetcode.com/problems/burst-balloons/
|
|
|
public class Code01_BurstBalloons {
|
|
|
|
|
|
public static int maxCoins0(int[] arr) {
|
|
|
// [3,2,1,3]
|
|
|
// [1,3,2,1,3,1]
|
|
|
int N = arr.length;
|
|
|
int[] help = new int[N + 2];
|
|
|
for (int i = 0; i < N; i++) {
|
|
|
help[i + 1] = arr[i];
|
|
|
}
|
|
|
help[0] = 1;
|
|
|
help[N + 1] = 1;
|
|
|
return func(help, 1, N);
|
|
|
}
|
|
|
|
|
|
// L-1位置，和R+1位置，永远不越界，并且，[L-1] 和 [R+1] 一定没爆呢！
|
|
|
// 返回，arr[L...R]打爆所有气球，最大得分是什么
|
|
|
public static int func(int[] arr, int L, int R) {
|
|
|
if (L == R) {
|
|
|
return arr[L - 1] * arr[L] * arr[R + 1];
|
|
|
}
|
|
|
// 尝试每一种情况，最后打爆的气球，是什么位置
|
|
|
// L...R
|
|
|
// L位置的气球，最后打爆
|
|
|
int max = func(arr, L + 1, R) + arr[L - 1] * arr[L] * arr[R + 1];
|
|
|
// R位置的气球，最后打爆
|
|
|
max = Math.max(max, func(arr, L, R - 1) + arr[L - 1] * arr[R] * arr[R + 1]);
|
|
|
// 尝试所有L...R，中间的位置，(L,R)
|
|
|
for (int i = L + 1; i < R; i++) {
|
|
|
// i位置的气球，最后打爆
|
|
|
int left = func(arr, L, i - 1);
|
|
|
int right = func(arr, i + 1, R);
|
|
|
int last = arr[L - 1] * arr[i] * arr[R + 1];
|
|
|
int cur = left + right + last;
|
|
|
max = Math.max(max, cur);
|
|
|
}
|
|
|
return max;
|
|
|
}
|
|
|
|
|
|
public static int maxCoins1(int[] arr) {
|
|
|
if (arr == null || arr.length == 0) {
|
|
|
return 0;
|
|
|
}
|
|
|
if (arr.length == 1) {
|
|
|
return arr[0];
|
|
|
}
|
|
|
int N = arr.length;
|
|
|
int[] help = new int[N + 2];
|
|
|
help[0] = 1;
|
|
|
help[N + 1] = 1;
|
|
|
for (int i = 0; i < N; i++) {
|
|
|
help[i + 1] = arr[i];
|
|
|
}
|
|
|
return process(help, 1, N);
|
|
|
}
|
|
|
|
|
|
// 打爆arr[L..R]范围上的所有气球，返回最大的分数
|
|
|
// 假设arr[L-1]和arr[R+1]一定没有被打爆
|
|
|
public static int process(int[] arr, int L, int R) {
|
|
|
if (L == R) {// 如果arr[L..R]范围上只有一个气球，直接打爆即可
|
|
|
return arr[L - 1] * arr[L] * arr[R + 1];
|
|
|
}
|
|
|
// 最后打爆arr[L]的方案，和最后打爆arr[R]的方案，先比较一下
|
|
|
int max = Math.max(arr[L - 1] * arr[L] * arr[R + 1] + process(arr, L + 1, R),
|
|
|
arr[L - 1] * arr[R] * arr[R + 1] + process(arr, L, R - 1));
|
|
|
// 尝试中间位置的气球最后被打爆的每一种方案
|
|
|
for (int i = L + 1; i < R; i++) {
|
|
|
max = Math.max(max, arr[L - 1] * arr[i] * arr[R + 1] + process(arr, L, i - 1) + process(arr, i + 1, R));
|
|
|
}
|
|
|
return max;
|
|
|
}
|
|
|
|
|
|
public static int maxCoins2(int[] arr) {
|
|
|
if (arr == null || arr.length == 0) {
|
|
|
return 0;
|
|
|
}
|
|
|
if (arr.length == 1) {
|
|
|
return arr[0];
|
|
|
}
|
|
|
int N = arr.length;
|
|
|
int[] help = new int[N + 2];
|
|
|
help[0] = 1;
|
|
|
help[N + 1] = 1;
|
|
|
for (int i = 0; i < N; i++) {
|
|
|
help[i + 1] = arr[i];
|
|
|
}
|
|
|
int[][] dp = new int[N + 2][N + 2];
|
|
|
for (int i = 1; i <= N; i++) {
|
|
|
dp[i][i] = help[i - 1] * help[i] * help[i + 1];
|
|
|
}
|
|
|
for (int L = N; L >= 1; L--) {
|
|
|
for (int R = L + 1; R <= N; R++) {
|
|
|
int ans = help[L - 1] * help[L] * help[R + 1] + dp[L + 1][R];
|
|
|
ans = Math.max(ans, help[L - 1] * help[R] * help[R + 1] + dp[L][R - 1]);
|
|
|
for (int i = L + 1; i < R; i++) {
|
|
|
ans = Math.max(ans, help[L - 1] * help[i] * help[R + 1] + dp[L][i - 1] + dp[i + 1][R]);
|
|
|
}
|
|
|
dp[L][R] = ans;
|
|
|
}
|
|
|
}
|
|
|
return dp[1][N];
|
|
|
}
|
|
|
|
|
|
}
|
