feat(MEDIUM): 2004_longestOnes

Author: wangpeng

src/pp/arithmetic/leetcode/_1004_longestOnes.java

@@ -0,0 +1,117 @@
+package pp.arithmetic.leetcode;
+
+/**
+ * Created by wangpeng on 2019-03-26.
+ * 1004. 最大连续1的个数 III
+ * <p>
+ * 给定一个由若干 0 和 1 组成的数组 A，我们最多可以将 K 个值从 0 变成 1 。
+ * <p>
+ * 返回仅包含 1 的最长（连续）子数组的长度。
+ * <p>
+ * <p>
+ * <p>
+ * 示例 1：
+ * <p>
+ * 输入：A = [1,1,1,0,0,0,1,1,1,1,0], K = 2
+ * 输出：6
+ * 解释：
+ * [1,1,1,0,0,1,1,1,1,1,1]
+ * 粗体数字从 0 翻转到 1，最长的子数组长度为 6。
+ * 示例 2：
+ * <p>
+ * 输入：A = [0,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,1], K = 3
+ * 输出：10
+ * 解释：
+ * [0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,1]
+ * 粗体数字从 0 翻转到 1，最长的子数组长度为 10。
+ * <p>
+ * <p>
+ * 提示：
+ * <p>
+ * 1 <= A.length <= 20000
+ * 0 <= K <= A.length
+ * A[i] 为 0 或 1
+ *
+ * @see <a href="https://leetcode-cn.com/problems/max-consecutive-ones-iii/">max-consecutive-ones-iii</a>
+ */
+public class _1004_longestOnes {
+
+    public static void main(String[] args) {
+        _1004_longestOnes ones = new _1004_longestOnes();
+        System.out.println(ones.longestOnes2(new int[]{1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0}, 2));
+        System.out.println(ones.longestOnes2(new int[]{0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1}, 3));
+        System.out.println(ones.longestOnes2(new int[]{1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1}, 144));
+    }
+
+    /**
+     * 解法优化，时间复杂度O(2n)
+     * 真正的是创建一个窗口进行移动，关键是在于窗口是如何移动的
+     *
+     * @param A
+     * @param K
+     * @return
+     */
+    public int longestOnes2(int[] A, int K) {
+        int lonest = 0;
+        int li = 0, ri = 0;
+        int zores = 0;
+        for (ri = 0; ri < A.length; ri++) {
+            if (A[ri] == 0) zores++;
+            while (zores > K) {
+                if (A[li++] == 0) zores--;
+            }
+            lonest = Math.max(lonest, ri - li + 1);
+        }
+
+        return lonest;
+    }
+
+    /**
+     * 耗时较严重，需要优化,时间复杂度O(n^2)
+     *
+     * @param A
+     * @param K
+     * @return
+     */
+    public int longestOnes(int[] A, int K) {
+        int left = -1, right = -1;
+        int count = K;
+        int longest = 0;
+        for (int i = 0; i < A.length; i++) {
+            //第一次都是从1开始
+            if (A[i] == 0 && left == right) {
+                continue;
+            }
+            if (left == right) {
+                left = i;
+                right = i + 1;
+                continue;
+            }
+            if (A[i] == 1) {
+                right = i;
+                continue;
+            }
+            if (A[i] == 0 && count > 0) {
+                right = i;
+                count--;
+                continue;
+            }
+            //消化结束
+            longest = Math.max(longest, right - left + 1);
+            //找下一个为0的位置开始
+            int index = left;
+            while (A[index] != 0 && index < A.length) {
+                index++;
+            }
+            i = index;
+            left = right = index;
+            count = K;
+        }
+        if (left != right) {
+            longest = Math.max(longest, right - left + 1 + count);
+        }
+        longest = Math.min(longest, A.length);
+
+        return longest;
+    }
+}