fd

Author: dingjikerbo

solution/src/main/java/com/inuker/solution/MaxConsecutiveOnesII.java

@@ -9,28 +9,42 @@
 
 public class MaxConsecutiveOnesII {
 
-    public int findMaxConsecutiveOnes(int[] nums) {
-        int max = 0, zero = 0, k = 1; // flip at most k zero
+    /**
+     * 这题思路是维护一个窗口[l,h]
+     * 这个窗口满足条件为0的个数小于等于k，这样这些0都能翻转成1，窗口的大小就是连续的1的个数了
+     * 如果遇到一个新的0，要判断当前窗口的0是否超了，如果超了，则窗口左边要收缩直到0的个数仍小于等于k为止
+     */
+    public int findMaxConsecutiveOnes(int[] nums, int k) {
+        int max = 0, zero = 0;
         for (int l = 0, h = 0; h < nums.length; h++) {
-            if (nums[h] == 0)
+            if (nums[h] == 0) {
                 zero++;
-            while (zero > k)
-                if (nums[l++] == 0)
+            }
+            while (zero > k) {
+                if (nums[l++] == 0) {
                     zero--;
+                }
+            }
             max = Math.max(max, h - l + 1);
         }
         return max;
     }
 
-    // for follow up
-    public int findMaxConsecutiveOnes2(int[] nums) {
-        int max = 0, k = 1; // flip at most k zero
+    /**
+     * for follow  up
+     * follow up的意思是说这是个数据流，不能像上面那样随机访问元素
+     * 这里用一个队列保存0元素的index，当新来一个0使得队列的size超过k时，队列要poll一个0
+     */
+    public int findMaxConsecutiveOnes2(int[] nums, int k) {
+        int max = 0;
         Queue<Integer> zeroIndex = new LinkedList<>();
         for (int l = 0, h = 0; h < nums.length; h++) {
-            if (nums[h] == 0)
+            if (nums[h] == 0) {
                 zeroIndex.offer(h);
-            if (zeroIndex.size() > k)
+            }
+            if (zeroIndex.size() > k) {
                 l = zeroIndex.poll() + 1;
+            }
             max = Math.max(max, h - l + 1);
         }
         return max;