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Author: sc-yan

Java/Array/11-Find Lucky Integer in an Array.java

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+Easy Leetcode 1394
+tags: Array
+
+方法1:
+题目不难。最直观的解法
+方法2：
+有点巧妙的解法。排序之后又是从后往前遍历
+
+方法3
+用array存出现的次数，而不是hashmap。这种情况index就是它的出现次数
+
+Approach 3A: Linked List (Beast Mode)
+
+The idea is to use the input array to store counts. This approach, however, is quite tricky to implement correctly!
+
+We can think of our array as a linked list, where arr[i] points to arr[arr[i] - 1] and so on, until the element that points to itself, or its outside of the array (and we do not care about that elements, per the intuition above).
+
+After we visit arr[arr[i] - 1], we can use that element to track the count of arr[i]. For the count, we will use a negative value to distinguish between pointers and counts.
+
+
+
+```
+
+/*
+方法1
+*/
+class Solution {
+    public int findLucky(int[] arr) {
+        HashMap<Integer, Integer> map = new HashMap<>();
+        for (int i = 0; i < arr.length; i++) {
+            map.put(arr[i], map.getOrDefault(arr[i], 0) +1);
+        }
+        int res = -1;
+        for (Map.Entry<Integer,Integer> x:map.entrySet()) {
+            if(x.getKey() == x.getValue()) {
+                res = Math.max(res, x.getValue());
+            }
+        }
+        return res;
+    }
+}
+
+/*
+方法2
+*/
+class Solution {
+    public int findLucky(int[] arr) {
+        Arrays.sort(arr);
+        int count = 1;
+        for (int i = arr.length-2; i >=0; i--) {
+            if(arr[i] == arr[i+1]) {
+                count++;
+            }else{
+                if(count == arr[i+1]){
+                    return count;
+                }
+                count = 1;
+            }
+        }
+        return arr[0] == count ? count :  -1;
+    }
+}
+/*
+方法3
+*/
+class Solution {
+    public int findLucky(int[] arr) {
+        int res = -1;
+        
+        if (arr == null || arr.length == 0)
+            return res;
+        
+        int[] store = new int[500+1]; // if we read the Constraints, the max length is 500
+        
+        for (int i = 0; i < arr.length; i++) {
+            store[arr[i]]++; // increment each elements in the store arr
+        }
+        
+        for (int i = store.length - 1; i >= 1; i--) { // iterate from the end
+            if (store[i] == i) // find the max element from the end, which equals condtions of the lucky integer
+                return store[i];
+        }
+        
+        return res;
+    }
+}
+    
+
+
+public static int findLucky(int[] arr) {
+        for(int i = 0; i < arr.length; i++) {
+            int j = i, tmp = arr[i];
+            while(tmp > 0 && tmp <= arr.length) {
+                int val = arr[tmp -1];
+                arr[tmp -1] = Math.min(0, arr[tmp -1]) -1;
+                if(tmp -1 <= i || tmp -1 == j) {
+                    break;
+                }
+                j = tmp -1;
+                tmp = val;
+            }
+        }
+        for(int i = arr.length; i > 0; i--) {
+            if(-arr[i -1] == i) {
+                return i;
+            }
+        }
+        return -1;
+    }