fd

Author: dingjikerbo

README.md

@@ -382,5 +382,6 @@
 |653|[Two Sum IV - Input is a BST](https://leetcode.com/problems/two-sum-iv-input-is-a-bst/)| [Java](leetcode/solution/src/TwoSumIV.java)|90||
 |654|[Maximum Binary Tree](https://leetcode.com/problems/maximum-binary-tree/description/)|[Java](leetcode/solution/src/MaximumBinaryTree.java)|100||
 |669|[Trim a Binary Search Tree](https://leetcode.com/problems/trim-a-binary-search-tree/)|[Java](leetcode/solution/src/TrimABinarySearchTree.java)|100||
+|681|[Next Closest Time](https://leetcode.com/problems/next-closest-time/)|[Java](leetcode/solution/src/NextClosestTime.java)|80|这题多做几遍|
 |684|[Redundant Connection](https://leetcode.com/problems/redundant-connection/)| [Java](leetcode/solution/src/RedundantConnection.java)|||
 |771|[Jewels and Stones](https://leetcode.com/problems/jewels-and-stones/)|[Java](leetcode/solution/src/JewelsAndStones.java)|100||

leetcode/solution/src/NextClosestTime.java

@@ -0,0 +1,62 @@
+import java.util.Arrays;
+import java.util.TreeSet;
+
+public class NextClosestTime {
+
+    /**
+     * 整体思路如下：
+     * 首先对输入的时间所包含的数要统计一下，并从小到大排序
+     * 接下来从时间的高位开始，寻找可替代的更大的数，如果找不到，就看下一位，如果找到了就替换成该数，判断时间是否有效
+     * 如果有效，则将该位往上全部替换为全局最小的数，并返回结果。否则还原该位，并继续看下一位。
+     */
+    public String nextClosestTime(String time) {
+        int[] number = new int[4];
+
+        time = time.substring(0, 2) + time.substring(3);
+
+        for (int i = 0; i < time.length(); i++) {
+            number[i] = time.charAt(i) - '0';
+        }
+
+        Arrays.sort(number);
+
+        StringBuilder sb = new StringBuilder(time);
+
+        int i;
+        for (i = 3; i >= 0; i--) {
+            char c = time.charAt(i);
+            int k = nextNumber(number, c - '0');
+            if (k >= 0) {
+                sb.setCharAt(i, (char) (k + '0'));
+                if (isValid(sb)) {
+                    break;
+                }
+                sb.setCharAt(i, c);
+            }
+        }
+        for (i++; i < 4; i++) {
+            sb.setCharAt(i, (char) (number[0] + '0'));
+        }
+        sb.insert(2, ':');
+        return sb.toString();
+    }
+
+    // 判断时间是否有效
+    private boolean isValid(StringBuilder sb) {
+        boolean flag1 = sb.charAt(0) < '2' || (sb.charAt(0) == '2' && sb.charAt(1) <= '3');
+        boolean flag2 = sb.charAt(2) < '5' || (sb.charAt(2) == '5' && sb.charAt(3) <= '9');
+        return flag1 && flag2;
+    }
+
+    /**
+     * 找比n更大的数，number是一个升序的数组
+     */
+    private int nextNumber(int[] number, int n) {
+        for (int i = 0; i < number.length; i++) {
+            if (number[i] > n) {
+                return number[i];
+            }
+        }
+        return -1;
+    }
+}

leetcode/src/Main.java

@@ -2,70 +2,55 @@
 
 public class Main {
 
-    public int search(int[] nums, int target) {
-        int left = 0, right = nums.length - 1;
+    public String nextClosestTime(String time) {
+        int[] number = new int[4];
 
-        while (left <= right) {
-            int mid = left + (right - left) / 2;
+        time = time.substring(0, 2) + time.substring(3);
 
-            if (nums[left] < nums[mid]) {
-                if (target >= nums[left] && target < nums[mid]) {
-                    right = mid - 1;
-                } else if (target == nums[mid]) {
-                    return mid;
-                } else {
-                    left = mid + 1;
-                }
-            } else if (nums[left] == nums[mid]) {
-                if (target == nums[mid]) {
-                    return mid;
-                } else {
-                    left++;
-                }
-            }
-            else {
-                if (target > nums[mid] && target <= nums[right]) {
-                    left = mid + 1;
-                } else if (target == nums[mid]) {
-                    return mid;
-                } else {
-                    right = mid - 1;
-                }
-            }
+        for (int i = 0; i < time.length(); i++) {
+            number[i] = time.charAt(i) - '0';
         }
 
-        return -1;
-    }
+        Arrays.sort(number);
 
-    // [3,1], 1
-    public int search2(int[] nums, int target) {
-        int left = 0, right = nums.length - 1;
+        StringBuilder sb = new StringBuilder(time);
 
-        while (left <= right) {
-            int mid = left + (right - left) / 2;
-
-            if (target == nums[mid]) {
-                return mid;
-            }
-
-            if (nums[left] <= nums[mid]) {
-                if (target >= nums[left] && target < nums[mid]) {
-                    right = mid - 1;
-                } else {
-                    left = mid + 1;
-                }
-            } else {
-                if (target > nums[mid] && target <= nums[right]) {
-                    left = mid + 1;
-                } else {
-                    right = mid - 1;
+        int i;
+        for (i = 3; i >= 0; i--) {
+            char c = time.charAt(i);
+            int k = nextNumber(number, c - '0');
+            if (k >= 0) {
+                sb.setCharAt(i, (char) (k + '0'));
+                if (isValid(sb)) {
+                    break;
                 }
+                sb.setCharAt(i, c);
             }
         }
+        for (i++; i < 4; i++) {
+            sb.setCharAt(i, (char) (number[0] + '0'));
+        }
+        sb.insert(2, ':');
+        return sb.toString();
+    }
 
+    private boolean isValid(StringBuilder sb) {
+        boolean flag1 = sb.charAt(0) < '2' || (sb.charAt(0) == '2' && sb.charAt(1) <= '3');
+        boolean flag2 = sb.charAt(2) < '5' || (sb.charAt(2) == '5' && sb.charAt(3) <= '9');
+        return flag1 && flag2;
+    }
+
+    private int nextNumber(int[] number, int n) {
+        for (int i = 0; i < number.length; i++) {
+            if (number[i] > n) {
+                return number[i];
+            }
+        }
         return -1;
     }
 
     public static void main(String[] args) {
+        String s = new NextClosestTime().nextClosestTime("13:55");
+        System.out.println(s);
     }
 }