算法习题之滑动窗口
滑动窗口
- 习题1 假设一个固定大小为W的窗口，依次划过arr，返回每一次滑出状况的最大值例 如，arr = [4,3,5,4,3,3,6,7], W = 3返回：[5,5,5,4,6,7]
- 习题2 给定一个整型数组arr，和一个整数num 某个arr中的子数组sub，如果想达标，必须满足：sub中最大值 – sub中最小值 <= num，返回arr中达标子数组的数量
- 习题3 加油站的良好出发点问题
- 习题4 arr是货币数组，其中的值都是正数。再给定一个正数aim。每个值都认为是一张货币，返回组成aim的最少货币数 注意：因为是求最少货币数，所以每一张货币认为是相同或者不同就不重要了
滑动窗口是什么？
滑动窗口是一种想象出来的数据结构：
滑动窗口有左边界L和有边界R
在数组或者字符串或者一个序列上，记为S，窗口就是S[L…R]这一部分
L往右滑意味着一个样本出了窗口，R往右滑意味着一个样本进了窗口
L和R都只能往右滑
滑动内最大值和最小值的更新结构
窗口不管L还是R滑动之后，都会让窗口呈现新状况，
如何能够更快的得到窗口当前状况下的最大值和最小值？
最好平均下来复杂度能做到O(1)
利用单调双端队列！
习题1 假设一个固定大小为W的窗口，依次划过arr，返回每一次滑出状况的最大值例 如，arr = [4,3,5,4,3,3,6,7], W = 3返回：[5,5,5,4,6,7]
// 暴力的对数器方法
public static int[] right(int[] arr, int w) {
if (arr == null || w < 1 || arr.length < w) {
return null;
}
int N = arr.length;
int[] res = new int[N - w + 1];
int index = 0;
int L = 0;
int R = w - 1;
while (R < N) {
int max = arr[L];
for (int i = L + 1; i <= R; i++) {
max = Math.max(max, arr[i]);
}
res[index++] = max;
L++;
R++;
}
return res;
}
public static int[] getMaxWindow(int[] arr, int w) {
if (arr == null || w < 1 || arr.length < w) {
return null;
}
// qmax 窗口最大值的更新结构
// 放下标
LinkedList qmax = new LinkedList();
int[] res = new int[arr.length - w + 1];
int index = 0;
for (int R = 0; R < arr.length; R++) {
while (!qmax.isEmpty() && arr[qmax.peekLast()] <= arr[R]) {
qmax.pollLast();
}
qmax.addLast(R);
if (qmax.peekFirst() == R - w) {
qmax.pollFirst();
}
if (R >= w - 1) {
res[index++] = arr[qmax.peekFirst()];
}
}
return res;
}
// for test
public static int[] generateRandomArray(int maxSize, int maxValue) {
int[] arr = new int[(int) ((maxSize + 1) * Math.random())];
for (int i = 0; i < arr.length; i++) {
arr[i] = (int) (Math.random() * (maxValue + 1));
}
return arr;
}
// for test
public static boolean isEqual(int[] arr1, int[] arr2) {
if ((arr1 == null && arr2 != null) || (arr1 != null && arr2 == null)) {
return false;
}
if (arr1 == null && arr2 == null) {
return true;
}
if (arr1.length != arr2.length) {
return false;
}
for (int i = 0; i < arr1.length; i++) {
if (arr1[i] != arr2[i]) {
return false;
}
}
return true;
}
public static void main(String[] args) {
int testTime = 100000;
int maxSize = 100;
int maxValue = 100;
System.out.println("test begin");
for (int i = 0; i < testTime; i++) {
int[] arr = generateRandomArray(maxSize, maxValue);
int w = (int) (Math.random() * (arr.length + 1));
int[] ans1 = getMaxWindow(arr, w);
int[] ans2 = right(arr, w);
if (!isEqual(ans1, ans2)) {
System.out.println("Oops!");
}
}
System.out.println("test finish");
}
习题2 给定一个整型数组arr，和一个整数num 某个arr中的子数组sub，如果想达标，必须满足：sub中最大值 – sub中最小值 <= num，返回arr中达标子数组的数量
// 暴力的对数器方法
public static int right(int[] arr, int sum) {
if (arr == null || arr.length == 0 || sum < 0) {
return 0;
}
int N = arr.length;
int count = 0;
for (int L = 0; L < N; L++) {
for (int R = L; R < N; R++) {
int max = arr[L];
int min = arr[L];
for (int i = L + 1; i <= R; i++) {
max = Math.max(max, arr[i]);
min = Math.min(min, arr[i]);
}
if (max - min <= sum) {
count++;
}
}
}
return count;
}
public static int num(int[] arr, int sum) {
if (arr == null || arr.length == 0 || sum < 0) {
return 0;
}
int N = arr.length;
int count = 0;
LinkedList maxWindow = new LinkedList<>();
LinkedList minWindow = new LinkedList<>();
int R = 0;
for (int L = 0; L < N; L++) {
while (R < N) {
while (!maxWindow.isEmpty() && arr[maxWindow.peekLast()] <= arr[R]) {
maxWindow.pollLast();
}
maxWindow.addLast(R);
while (!minWindow.isEmpty() && arr[minWindow.peekLast()] >= arr[R]) {
minWindow.pollLast();
}
minWindow.addLast(R);
if (arr[maxWindow.peekFirst()] - arr[minWindow.peekFirst()] > sum) {
break;
} else {
R++;
}
}
count += R - L;
if (maxWindow.peekFirst() == L) {
maxWindow.pollFirst();
}
if (minWindow.peekFirst() == L) {
minWindow.pollFirst();
}
}
return count;
}
// for test
public static int[] generateRandomArray(int maxLen, int maxValue) {
int len = (int) (Math.random() * (maxLen + 1));
int[] arr = new int[len];
for (int i = 0; i < len; i++) {
arr[i] = (int) (Math.random() * (maxValue + 1)) - (int) (Math.random() * (maxValue + 1));
}
return arr;
}
// for test
public static void printArray(int[] arr) {
if (arr != null) {
for (int i = 0; i < arr.length; i++) {
System.out.print(arr[i] + " ");
}
System.out.println();
}
}
public static void main(String[] args) {
int maxLen = 100;
int maxValue = 200;
int testTime = 100000;
System.out.println("测试开始");
for (int i = 0; i < testTime; i++) {
int[] arr = generateRandomArray(maxLen, maxValue);
int sum = (int) (Math.random() * (maxValue + 1));
int ans1 = right(arr, sum);
int ans2 = num(arr, sum);
if (ans1 != ans2) {
System.out.println("Oops!");
printArray(arr);
System.out.println(sum);
System.out.println(ans1);
System.out.println(ans2);
break;
}
}
System.out.println("测试结束");
}
习题3 加油站的良好出发点问题
// 这个方法的时间复杂度O(N)，额外空间复杂度O(N)
public static int canCompleteCircuit(int[] gas, int[] cost) {
boolean[] good = goodArray(gas, cost);
for (int i = 0; i < gas.length; i++) {
if (good[i]) {
return i;
}
}
return -1;
}
public static boolean[] goodArray(int[] g, int[] c) {
int N = g.length;
int M = N << 1;
int[] arr = new int[M];
for (int i = 0; i < N; i++) {
arr[i] = g[i] - c[i];
arr[i + N] = g[i] - c[i];
}
for (int i = 1; i < M; i++) {
arr[i] += arr[i - 1];
}
LinkedList w = new LinkedList<>();
for (int i = 0; i < N; i++) {
while (!w.isEmpty() && arr[w.peekLast()] >= arr[i]) {
w.pollLast();
}
w.addLast(i);
}
boolean[] ans = new boolean[N];
for (int offset = 0, i = 0, j = N; j < M; offset = arr[i++], j++) {
if (arr[w.peekFirst()] - offset >= 0) {
ans[i] = true;
}
if (w.peekFirst() == i) {
w.pollFirst();
}
while (!w.isEmpty() && arr[w.peekLast()] >= arr[j]) {
w.pollLast();
}
w.addLast(j);
}
return ans;
}
习题4 arr是货币数组，其中的值都是正数。再给定一个正数aim。每个值都认为是一张货币，返回组成aim的最少货币数 注意：因为是求最少货币数，所以每一张货币认为是相同或者不同就不重要了
public static int minCoins(int[] arr, int aim) {
return process(arr, 0, aim);
}
public static int process(int[] arr, int index, int rest) {
if (rest < 0) {
return Integer.MAX_VALUE;
}
if (index == arr.length) {
return rest == 0 ? 0 : Integer.MAX_VALUE;
} else {
int p1 = process(arr, index + 1, rest);
int p2 = process(arr, index + 1, rest - arr[index]);
if (p2 != Integer.MAX_VALUE) {
p2++;
}
return Math.min(p1, p2);
}
}
// dp1时间复杂度为：O(arr长度 * aim)
public static int dp1(int[] arr, int aim) {
if (aim == 0) {
return 0;
}
int N = arr.length;
int[][] dp = new int[N + 1][aim + 1];
dp[N][0] = 0;
for (int j = 1; j <= aim; j++) {
dp[N][j] = Integer.MAX_VALUE;
}
for (int index = N - 1; index >= 0; index--) {
for (int rest = 0; rest <= aim; rest++) {
int p1 = dp[index + 1][rest];
int p2 = rest - arr[index] >= 0 ? dp[index + 1][rest - arr[index]] : Integer.MAX_VALUE;
if (p2 != Integer.MAX_VALUE) {
p2++;
}
dp[index][rest] = Math.min(p1, p2);
}
}
return dp[0][aim];
}
public static class Info {
public int[] coins;
public int[] zhangs;
public Info(int[] c, int[] z) {
coins = c;
zhangs = z;
}
}
public static Info getInfo(int[] arr) {
HashMap counts = new HashMap<>();
for (int value : arr) {
if (!counts.containsKey(value)) {
counts.put(value, 1);
} else {
counts.put(value, counts.get(value) + 1);
}
}
int N = counts.size();
int[] coins = new int[N];
int[] zhangs = new int[N];
int index = 0;
for (Entry entry : counts.entrySet()) {
coins[index] = entry.getKey();
zhangs[index++] = entry.getValue();
}
return new Info(coins, zhangs);
}
// dp2时间复杂度为：O(arr长度) + O(货币种数 * aim * 每种货币的平均张数)
public static int dp2(int[] arr, int aim) {
if (aim == 0) {
return 0;
}
// 得到info时间复杂度O(arr长度)
Info info = getInfo(arr);
int[] coins = info.coins;
int[] zhangs = info.zhangs;
int N = coins.length;
int[][] dp = new int[N + 1][aim + 1];
dp[N][0] = 0;
for (int j = 1; j <= aim; j++) {
dp[N][j] = Integer.MAX_VALUE;
}
// 这三层for循环，时间复杂度为O(货币种数 * aim * 每种货币的平均张数)
for (int index = N - 1; index >= 0; index--) {
for (int rest = 0; rest <= aim; rest++) {
dp[index][rest] = dp[index + 1][rest];
for (int zhang = 1; zhang * coins[index] <= aim && zhang <= zhangs[index]; zhang++) {
if (rest - zhang * coins[index] >= 0
&& dp[index + 1][rest - zhang * coins[index]] != Integer.MAX_VALUE) {
dp[index][rest] = Math.min(dp[index][rest], zhang + dp[index + 1][rest - zhang * coins[index]]);
}
}
}
}
return dp[0][aim];
}
// dp3时间复杂度为：O(arr长度) + O(货币种数 * aim)
// 优化需要用到窗口内最小值的更新结构
public static int dp3(int[] arr, int aim) {
if (aim == 0) {
return 0;
}
// 得到info时间复杂度O(arr长度)
Info info = getInfo(arr);
int[] c = info.coins;
int[] z = info.zhangs;
int N = c.length;
int[][] dp = new int[N + 1][aim + 1];
dp[N][0] = 0;
for (int j = 1; j <= aim; j++) {
dp[N][j] = Integer.MAX_VALUE;
}
// 虽然是嵌套了很多循环，但是时间复杂度为O(货币种数 * aim)
// 因为用了窗口内最小值的更新结构
for (int i = N - 1; i >= 0; i--) {
for (int mod = 0; mod < Math.min(aim + 1, c[i]); mod++) {
// 当前面值 X
// mod mod + x mod + 2*x mod + 3 * x
LinkedList w = new LinkedList<>();
w.add(mod);
dp[i][mod] = dp[i + 1][mod];
for (int r = mod + c[i]; r <= aim; r += c[i]) {
while (!w.isEmpty() && (dp[i + 1][w.peekLast()] == Integer.MAX_VALUE
|| dp[i + 1][w.peekLast()] + compensate(w.peekLast(), r, c[i]) >= dp[i + 1][r])) {
w.pollLast();
}
w.addLast(r);
int overdue = r - c[i] * (z[i] + 1);
if (w.peekFirst() == overdue) {
w.pollFirst();
}
dp[i][r] = dp[i + 1][w.peekFirst()] + compensate(w.peekFirst(), r, c[i]);
}
}
}
return dp[0][aim];
}
public static int compensate(int pre, int cur, int coin) {
return (cur - pre) / coin;
}
// 为了测试
public static int[] randomArray(int N, int maxValue) {
int[] arr = new int[N];
for (int i = 0; i < N; i++) {
arr[i] = (int) (Math.random() * maxValue) + 1;
}
return arr;
}
// 为了测试
public static void printArray(int[] arr) {
for (int i = 0; i < arr.length; i++) {
System.out.print(arr[i] + " ");
}
System.out.println();
}
// 为了测试
public static void main(String[] args) {
int maxLen = 20;
int maxValue = 30;
int testTime = 300000;
System.out.println("功能测试开始");
for (int i = 0; i < testTime; i++) {
int N = (int) (Math.random() * maxLen);
int[] arr = randomArray(N, maxValue);
int aim = (int) (Math.random() * maxValue);
int ans1 = minCoins(arr, aim);
int ans2 = dp1(arr, aim);
int ans3 = dp2(arr, aim);
int ans4 = dp3(arr, aim);
if (ans1 != ans2 || ans3 != ans4 || ans1 != ans3) {
System.out.println("Oops!");
printArray(arr);
System.out.println(aim);
System.out.println(ans1);
System.out.println(ans2);
System.out.println(ans3);
System.out.println(ans4);
break;
}
}
System.out.println("功能测试结束");
System.out.println("==========");
int aim = 0;
int[] arr = null;
long start;
long end;
int ans2;
int ans3;
System.out.println("性能测试开始");
maxLen = 30000;
maxValue = 20;
aim = 60000;
arr = randomArray(maxLen, maxValue);
start = System.currentTimeMillis();
ans2 = dp2(arr, aim);
end = System.currentTimeMillis();
System.out.println("dp2答案 : " + ans2 + ", dp2运行时间 : " + (end - start) + " ms");
start = System.currentTimeMillis();
ans3 = dp3(arr, aim);
end = System.currentTimeMillis();
System.out.println("dp3答案 : " + ans3 + ", dp3运行时间 : " + (end - start) + " ms");
System.out.println("性能测试结束");
System.out.println("===========");
System.out.println("货币大量重复出现情况下，");
System.out.println("大数据量测试dp3开始");
maxLen = 20000000;
aim = 10000;
maxValue = 10000;
arr = randomArray(maxLen, maxValue);
start = System.currentTimeMillis();
ans3 = dp3(arr, aim);
end = System.currentTimeMillis();
System.out.println("dp3运行时间 : " + (end - start) + " ms");
System.out.println("大数据量测试dp3结束");
System.out.println("===========");
System.out.println("当货币很少出现重复，dp2比dp3有常数时间优势");
System.out.println("当货币大量出现重复，dp3时间复杂度明显优于dp2");
System.out.println("dp3的优化用到了窗口内最小值的更新结构");
}