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modify code

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pull/6/head
左程云 5 years ago
parent 10f70fbe68
commit 8f8a26a40a
3 changed files with 167 additions and 105 deletions
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46
src/class20/Code01_PalindromeSubsequence.java
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@ -3,6 +3,52 @@ package class20;
// 测试链接：https://leetcode.com/problems/longest-palindromic-subsequence/
public class Code01_PalindromeSubsequence {
public static int lpsl1(String s) {
if (s == null || s.length() == 0) {
return 0;
}
char[] str = s.toCharArray();
return f(str, 0, str.length - 1);
}
// str[L..R]最长回文子序列长度返回
public static int f(char[] str, int L, int R) {
if (L == R) {
return 1;
}
if (L == R - 1) {
return str[L] == str[R] ? 2 : 1;
}
int p1 = f(str, L + 1, R - 1);
int p2 = f(str, L, R - 1);
int p3 = f(str, L + 1, R);
int p4 = str[L] != str[R] ? 0 : (2 + f(str, L + 1, R - 1));
return Math.max(Math.max(p1, p2), Math.max(p3, p4));
}
public static int lpsl2(String s) {
if (s == null || s.length() == 0) {
return 0;
}
char[] str = s.toCharArray();
int N = str.length;
int[][] dp = new int[N][N];
dp[N - 1][N - 1] = 1;
for (int i = 0; i < N - 1; i++) {
dp[i][i] = 1;
dp[i][i + 1] = str[i] == str[i + 1] ? 2 : 1;
}
for (int L = N - 3; L >= 0; L--) {
for (int R = L + 2; R < N; R++) {
dp[L][R] = Math.max(dp[L][R - 1], dp[L + 1][R]);
if (str[L] == str[R]) {
dp[L][R] = Math.max(dp[L][R], 2 + dp[L + 1][R - 1]);
}
}
}
return dp[0][N - 1];
}
public static int longestPalindromeSubseq1(String s) {
if (s == null || s.length() == 0) {
return 0;

90
src/class20/Code02_HorseJump.java
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@ -2,6 +2,60 @@ package class20;
public class Code02_HorseJump {
// 当前来到的位置是（x,y）
// 还剩下rest步需要跳
// 跳完rest步，正好跳到a，b的方法数是多少？
// 10 * 9
public static int jump(int a, int b, int k) {
return process(0, 0, k, a, b);
}
public static int process(int x, int y, int rest, int a, int b) {
if (x < 0 || x > 9 || y < 0 || y > 8) {
return 0;
}
if (rest == 0) {
return (x == a && y == b) ? 1 : 0;
}
int ways = process(x + 2, y + 1, rest - 1, a, b);
ways += process(x + 1, y + 2, rest - 1, a, b);
ways += process(x - 1, y + 2, rest - 1, a, b);
ways += process(x - 2, y + 1, rest - 1, a, b);
ways += process(x - 2, y - 1, rest - 1, a, b);
ways += process(x - 1, y - 2, rest - 1, a, b);
ways += process(x + 1, y - 2, rest - 1, a, b);
ways += process(x + 2, y - 1, rest - 1, a, b);
return ways;
}
public static int dp(int a, int b, int k) {
int[][][] dp = new int[10][9][k + 1];
dp[a][b][0] = 1;
for (int rest = 1; rest <= k; rest++) {
for (int x = 0; x < 10; x++) {
for (int y = 0; y < 9; y++) {
int ways = pick(dp, x + 2, y + 1, rest - 1);
ways += pick(dp, x + 1, y + 2, rest - 1);
ways += pick(dp, x - 1, y + 2, rest - 1);
ways += pick(dp, x - 2, y + 1, rest - 1);
ways += pick(dp, x - 2, y - 1, rest - 1);
ways += pick(dp, x - 1, y - 2, rest - 1);
ways += pick(dp, x + 1, y - 2, rest - 1);
ways += pick(dp, x + 2, y - 1, rest - 1);
dp[x][y][rest] = ways;
}
}
}
return dp[0][0][k];
}
public static int pick(int[][][] dp, int x, int y, int rest) {
if (x < 0 || x > 9 || y < 0 || y > 8) {
return 0;
}
return dp[x][y][rest];
}
public static int ways(int a, int b, int step) {
return f(0, 0, step, a, b);
}
@ -13,32 +67,22 @@ public class Code02_HorseJump {
if (step == 0) {
return (i == a && j == b) ? 1 : 0;
}
return f(i - 2, j + 1, step - 1, a, b)
+ f(i - 1, j + 2, step - 1, a, b)
+ f(i + 1, j + 2, step - 1, a, b)
+ f(i + 2, j + 1, step - 1, a, b)
+ f(i + 2, j - 1, step - 1, a, b)
+ f(i + 1, j - 2, step - 1, a, b)
+ f(i - 1, j - 2, step - 1, a, b)
+ f(i - 2, j - 1, step - 1, a, b);
return f(i - 2, j + 1, step - 1, a, b) + f(i - 1, j + 2, step - 1, a, b) + f(i + 1, j + 2, step - 1, a, b)
+ f(i + 2, j + 1, step - 1, a, b) + f(i + 2, j - 1, step - 1, a, b) + f(i + 1, j - 2, step - 1, a, b)
+ f(i - 1, j - 2, step - 1, a, b) + f(i - 2, j - 1, step - 1, a, b);
}
public static int waysdp(int a, int b, int s) {
int[][][] dp = new int[10][9][s+1];
int[][][] dp = new int[10][9][s + 1];
dp[a][b][0] = 1;
for(int step = 1 ; step <= s;step++ ) { // 按层来
for(int i = 0 ; i < 10;i++) {
for(int j = 0 ; j < 9; j++) {
dp[i][j][step] = getValue(dp,i - 2, j + 1, step - 1)
+ getValue(dp,i - 1, j + 2, step - 1)
+ getValue(dp,i + 1, j + 2, step - 1)
+ getValue(dp,i + 2, j + 1, step - 1)
+ getValue(dp,i + 2, j - 1, step - 1)
+ getValue(dp,i + 1, j - 2, step - 1)
+ getValue(dp,i - 1, j - 2, step - 1)
+ getValue(dp,i - 2, j - 1, step - 1);
for (int step = 1; step <= s; step++) { // 按层来
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 9; j++) {
dp[i][j][step] = getValue(dp, i - 2, j + 1, step - 1) + getValue(dp, i - 1, j + 2, step - 1)
+ getValue(dp, i + 1, j + 2, step - 1) + getValue(dp, i + 2, j + 1, step - 1)
+ getValue(dp, i + 2, j - 1, step - 1) + getValue(dp, i + 1, j - 2, step - 1)
+ getValue(dp, i - 1, j - 2, step - 1) + getValue(dp, i - 2, j - 1, step - 1);
}
}
}
@ -58,6 +102,8 @@ public class Code02_HorseJump {
int y = 7;
int step = 10;
System.out.println(ways(x, y, step));
System.out.println(waysdp(x, y, step));
System.out.println(dp(x, y, step));
System.out.println(jump(x, y, step));
}
}

134
src/class20/Code03_Coffee.java
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@ -14,14 +14,14 @@ import java.util.PriorityQueue;
// 假设时间点从0开始，返回所有人喝完咖啡并洗完咖啡杯的全部过程结束后，至少来到什么时间点。
public class Code03_Coffee {
// 方法一：暴力尝试方法
public static int minTime1(int[] arr, int n, int a, int b) {
// 验证的方法，彻底的暴力，不要脸的暴力
public static int right(int[] arr, int n, int a, int b) {
int[] times = new int[arr.length];
int[] drink = new int[n];
return forceMake(arr, times, 0, drink, n, a, b);
}
// 方法一，每个人暴力尝试用每一个咖啡机给自己做咖啡
// 每个人暴力尝试用每一个咖啡机给自己做咖啡
public static int forceMake(int[] arr, int[] times, int kth, int[] drink, int n, int a, int b) {
if (kth == n) {
int[] drinkSorted = Arrays.copyOf(drink, kth);
@ -41,7 +41,6 @@ public class Code03_Coffee {
return time;
}
// 方法一，暴力尝试洗咖啡杯的方式
public static int forceWash(int[] drinks, int a, int b, int index, int washLine, int time) {
if (index == drinks.length) {
return time;
@ -56,7 +55,7 @@ public class Code03_Coffee {
return Math.min(ans1, ans2);
}
// 方法二：稍微好一点的解法
// 以下为贪心+优良暴力
public static class Machine {
public int timePoint;
public int workTime;
@ -76,8 +75,8 @@ public class Code03_Coffee {
}
// 方法二，每个人暴力尝试用每一个咖啡机给自己做咖啡，优化成贪心
public static int minTime2(int[] arr, int n, int a, int b) {
// 优良一点的暴力尝试的方法
public static int minTime1(int[] arr, int n, int a, int b) {
PriorityQueue<Machine> heap = new PriorityQueue<Machine>(new MachineComparator());
for (int i = 0; i < arr.length; i++) {
heap.add(new Machine(0, arr[i]));
@ -89,66 +88,32 @@ public class Code03_Coffee {
drinks[i] = cur.timePoint;
heap.add(cur);
}
return process(drinks, a, b, 0, 0);
}
// 方法二，洗咖啡杯的方式和原来一样，只是这个暴力版本减少了一个可变参数
// process(drinks, 3, 10, 0,0)
// a 洗一杯的时间 固定变量
// b 自己挥发干净的时间 固定变量
// drinks 每一个员工喝完的时间 固定变量
// drinks[0..index-1]都已经干净了，不用你操心了
// drinks[index...]都想变干净，这是我操心的，washLine表示洗的机器何时可用
// drinks[index...]变干净，最少的时间点返回
public static int process(int[] drinks, int a, int b, int index, int washLine) {
if (index == drinks.length - 1) {
return Math.min(Math.max(washLine, drinks[index]) + a, drinks[index] + b);
}
// 剩不止一杯咖啡
// wash是我当前的咖啡杯，洗完的时间
int wash = Math.max(washLine, drinks[index]) + a;// 洗，index一杯，结束的时间点
// index+1...变干净的最早时间
int next1 = process(drinks, a, b, index + 1, wash);
// index....
int p1 = Math.max(wash, next1);
int dry = drinks[index] + b; // 挥发，index一杯，结束的时间点
int next2 = process(drinks, a, b, index + 1, washLine);
int p2 = Math.max(dry, next2);
return Math.min(p1, p2);
return bestTime(drinks, a, b, 0, 0);
}
public static int dp(int[] drinks, int a, int b) {
if (a >= b) {
return drinks[drinks.length - 1] + b;
}
// a < b
int N = drinks.length;
int limit = 0; // 咖啡机什么时候可用
for (int i = 0; i < N; i++) {
limit = Math.max(limit, drinks[i]) + a;
}
int[][] dp = new int[N][limit + 1];
// N-1行，所有的值
for (int washLine = 0; washLine <= limit; washLine++) {
dp[N - 1][washLine] = Math.min(Math.max(washLine, drinks[N - 1]) + a, drinks[N - 1] + b);
}
for (int index = N - 2; index >= 0; index--) {
for (int washLine = 0; washLine <= limit; washLine++) {
int p1 = Integer.MAX_VALUE;
int wash = Math.max(washLine, drinks[index]) + a;
if (wash <= limit) {
p1 = Math.max(wash, dp[index + 1][wash]);
}
int p2 = Math.max(drinks[index] + b, dp[index + 1][washLine]);
dp[index][washLine] = Math.min(p1, p2);
}
// drinks 所有杯子可以开始洗的时间
// wash 单杯洗干净的时间（串行）
// air 挥发干净的时间(并行)
// free 洗的机器什么时候可用
// drinks[index.....]都变干净，最早的结束时间（返回）
public static int bestTime(int[] drinks, int wash, int air, int index, int free) {
if (index == drinks.length) {
return 0;
}
return dp[0][0];
// index号杯子 决定洗
int selfClean1 = Math.max(drinks[index], free) + wash;
int restClean1 = bestTime(drinks, wash, air, index + 1, selfClean1);
int p1 = Math.max(selfClean1, restClean1);
// index号杯子 决定挥发
int selfClean2 = drinks[index] + air;
int restClean2 = bestTime(drinks, wash, air, index + 1, free);
int p2 = Math.max(selfClean2, restClean2);
return Math.min(p1, p2);
}
// 方法三：最终版本，把方法二洗咖啡杯的暴力尝试进一步优化成动态规划
public static int minTime3(int[] arr, int n, int a, int b) {
// 方法二，贪心+优良尝试改成动态规划
public static int minTime2(int[] arr, int n, int a, int b) {
PriorityQueue<Machine> heap = new PriorityQueue<Machine>(new MachineComparator());
for (int i = 0; i < arr.length; i++) {
heap.add(new Machine(0, arr[i]));
@ -160,18 +125,29 @@ public class Code03_Coffee {
drinks[i] = cur.timePoint;
heap.add(cur);
}
if (a >= b) {
return drinks[n - 1] + b;
return bestTimeDp(drinks, a, b);
}
int[][] dp = new int[n][drinks[n - 1] + n * a];
for (int i = 0; i < dp[0].length; i++) {
dp[n - 1][i] = Math.min(Math.max(i, drinks[n - 1]) + a, drinks[n - 1] + b);
}
for (int row = n - 2; row >= 0; row--) { // row 咖啡杯的编号
int washLine = drinks[row] + (row + 1) * a;
for (int col = 0; col < washLine; col++) {
int wash = Math.max(col, drinks[row]) + a;
dp[row][col] = Math.min(Math.max(wash, dp[row + 1][wash]), Math.max(drinks[row] + b, dp[row + 1][col]));
public static int bestTimeDp(int[] drinks, int wash, int air) {
int N = drinks.length;
int maxFree = 0;
for (int i = 0; i < drinks.length; i++) {
maxFree = Math.max(maxFree, drinks[i]) + wash;
}
int[][] dp = new int[N + 1][maxFree + 1];
for (int index = N - 1; index >= 0; index--) {
for (int free = 0; free <= maxFree; free++) {
int selfClean1 = Math.max(drinks[index], free) + wash;
if (selfClean1 > maxFree) {
break; // 因为free后面的也都不用填了
}
int restClean1 = dp[index + 1][selfClean1];
int p1 = Math.max(selfClean1, restClean1);
// index号杯子 决定挥发
int selfClean2 = drinks[index] + air;
int restClean2 = dp[index + 1][free];
int p2 = Math.max(selfClean2, restClean2);
dp[index][free] = Math.min(p1, p2);
}
}
return dp[0][0];
@ -196,12 +172,6 @@ public class Code03_Coffee {
}
public static void main(String[] args) {
int[] test = { 1, 1, 5, 5, 7, 10, 12, 12, 12, 12, 12, 12, 15 };
int a1 = 3;
int b1 = 10;
System.out.println(process(test, a1, b1, 0, 0));
System.out.println(dp(test, a1, b1));
int len = 5;
int max = 9;
int testTime = 50000;
@ -210,9 +180,9 @@ public class Code03_Coffee {
int n = (int) (Math.random() * 5) + 1;
int a = (int) (Math.random() * 5) + 1;
int b = (int) (Math.random() * 10) + 1;
int ans1 = minTime1(arr, n, a, b);
int ans2 = minTime2(arr, n, a, b);
int ans3 = minTime3(arr, n, a, b);
int ans1 = right(arr, n, a, b);
int ans2 = minTime1(arr, n, a, b);
int ans3 = minTime2(arr, n, a, b);
if (ans1 != ans2 || ans2 != ans3) {
printArray(arr);
System.out.println("n : " + n);

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