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package class14;
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import java.util.HashSet;
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public class Code01_Light {
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public static int minLight1(String road) {
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if (road == null || road.length() == 0) {
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return 0;
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}
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return process(road.toCharArray(), 0, new HashSet<>());
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}
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// str[index....]位置，自由选择放灯还是不放灯
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// str[0..index-1]位置呢？已经做完决定了，那些放了灯的位置，存在lights里
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// 要求选出能照亮所有.的方案，并且在这些有效的方案中，返回最少需要几个灯
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public static int process(char[] str, int index, HashSet<Integer> lights) {
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if (index == str.length) { // 结束的时候
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for (int i = 0; i < str.length; i++) {
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if (str[i] != 'X') { // 当前位置是点的话
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if (!lights.contains(i - 1) && !lights.contains(i) && !lights.contains(i + 1)) {
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return Integer.MAX_VALUE;
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}
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}
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}
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return lights.size();
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} else { // str还没结束
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// i X .
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int no = process(str, index + 1, lights);
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int yes = Integer.MAX_VALUE;
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if (str[index] == '.') {
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lights.add(index);
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yes = process(str, index + 1, lights);
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lights.remove(index);
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}
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return Math.min(no, yes);
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}
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}
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public static int minLight2(String road) {
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char[] str = road.toCharArray();
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int i = 0;
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int light = 0;
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while (i < str.length) {
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if (str[i] == 'X') {
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i++;
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} else {
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light++;
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if (i + 1 == str.length) {
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break;
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} else { // 有i位置 i+ 1 X .
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if (str[i + 1] == 'X') {
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i = i + 2;
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} else {
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i = i + 3;
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}
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}
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}
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}
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return light;
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}
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// 更简洁的解法
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// 两个X之间，数一下.的数量，然后除以3，向上取整
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// 把灯数累加
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public static int minLight3(String road) {
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char[] str = road.toCharArray();
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int cur = 0;
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int light = 0;
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for (char c : str) {
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if (c == 'X') {
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light += (cur + 2) / 3;
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cur = 0;
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} else {
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cur++;
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}
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}
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light += (cur + 2) / 3;
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return light;
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}
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// for test
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public static String randomString(int len) {
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char[] res = new char[(int) (Math.random() * len) + 1];
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for (int i = 0; i < res.length; i++) {
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res[i] = Math.random() < 0.5 ? 'X' : '.';
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}
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return String.valueOf(res);
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}
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public static void main(String[] args) {
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int len = 20;
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int testTime = 100000;
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for (int i = 0; i < testTime; i++) {
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String test = randomString(len);
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int ans1 = minLight1(test);
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int ans2 = minLight2(test);
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int ans3 = minLight3(test);
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if (ans1 != ans2 || ans1 != ans3) {
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System.out.println("oops!");
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}
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}
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System.out.println("finish!");
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}
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}
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