LeetCode第605题 - 种花问题

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] == 0
            counts[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));
}