题目
解答
方案一
public class Solution {public boolean canPlaceFlowers(int[] flowerbed, int n) {if (flowerbed == null || flowerbed.length == 0) {return false;}if (flowerbed.length < 2) {return (flowerbed[0] == 1 ? 0 : 1) >= n;}if (flowerbed.length < 3) {int count = 0;if (flowerbed[0] == 0 && flowerbed[1] == 0) {count = 1;}return count >= n;}int length = flowerbed.length;int[] counts = new int[length];if (flowerbed[0] == 0 && flowerbed[1] == 0) {counts[0] = 1;}if (flowerbed[0] == 1) {counts[1] = 0;} else {// flowerbed[0] == 0counts[1] = flowerbed[1] == 1 ? 0 : 1;}int max = length - 1;for (int i = 2; i < max; ++i) {if (flowerbed[i - 1] == 1 || flowerbed[i] == 1 || flowerbed[i + 1] == 1) {counts[i] = counts[i - 1];} else {counts[i] = Math.max(counts[i - 2] + 1, counts[i - 1]);}if (counts[i] >= n) {return true;}}if (flowerbed[max] == 1 || flowerbed[max - 1] == 1) {counts[max] = counts[max - 1];} else {counts[max] = Math.max(counts[max - 2] + 1, counts[max - 1]);}return counts[max] >= n;}
}
方案二
public class Solution {public boolean canPlaceFlowers(int[] flowerbed, int n) {int length = flowerbed.length;int[] flowerbedNew = new int[length + 2];System.arraycopy(flowerbed, 0, flowerbedNew, 1, length);flowerbedNew[0] = 0;flowerbedNew[length + 1] = 0;for (int i = 1; i < length; ++i) {if (flowerbedNew[i - 1] == 0 && flowerbedNew[i] == 0 && flowerbedNew[i + 1] == 0) {--n;flowerbedNew[i] = 1;}if (n <= 0) {return true;}}if (n <= 0) {return true;}return false;}
}
要点
直观的说，方案一的实现很丑陋。
依据题目的定义，有个简单的思路，即每三个位置可以种一株花，那么在输入数据的前、后增加哨兵元素，可以规避边界条件的处理，一定程度上简化实现。
本题目在贪心算法的分类下，后续补充基于贪心算法的求解实现。
准备的用例，如下
@Before
public void before() {t = new Solution();
}@Test
public void test01() {assertTrue(t.canPlaceFlowers(new int[] { 1, 0, 0, 0, 1 }, 1));
}@Test
public void test02() {assertFalse(t.canPlaceFlowers(new int[] { 1, 0, 0, 0, 1 }, 2));
}@Test
public void test03() {assertFalse(t.canPlaceFlowers(new int[] { 1 }, 2));
}@Test
public void test04() {assertFalse(t.canPlaceFlowers(new int[] { 1 }, 1));
}@Test
public void test05() {assertTrue(t.canPlaceFlowers(new int[] { 1 }, 0));
}@Test
public void test06() {assertTrue(t.canPlaceFlowers(new int[] { 0 }, 0));
}@Test
public void test07() {assertFalse(t.canPlaceFlowers(new int[] { 1, 0 }, 2));
}@Test
public void test08() {assertFalse(t.canPlaceFlowers(new int[] { 1, 0 }, 1));
}@Test
public void test09() {assertTrue(t.canPlaceFlowers(new int[] { 0, 0 }, 1));
}@Test
public void test10() {assertFalse(t.canPlaceFlowers(new int[] { 0, 0 }, 2));
}@Test
public void test11() {assertFalse(t.canPlaceFlowers(new int[] { 1, 0, 0, 0, 0, 1 }, 2));
}@Test
public void test12() {assertTrue(t.canPlaceFlowers(new int[] { 0, 0, 1, 0, 1 }, 1));
}@Test
public void test13() {assertFalse(t.canPlaceFlowers(new int[] { 0, 1, 0 }, 1));
}@Test
public void test14() {assertFalse(t.canPlaceFlowers(new int[] { 1, 0, 1, 0, 0, 1, 0 }, 1));
}
参数详解)
