Tuesday, February 7, 2017

LeetCode 496 - Next Greater Element I


LeetCode 496 - Next Greater Element I
LeetCode 503 - Next Greater Element II
https://leetcode.com/problems/next-greater-element-i/
You are given two arrays (without duplicates) nums1 and nums2 where nums1’s elements are subset of nums2. Find all the next greater numbers for nums1's elements in the corresponding places of nums2.
The Next Greater Number of a number x in nums1 is the first greater number to its right in nums2. If it does not exist, output -1 for this number.
Example 1:
Input: nums1 = [4,1,2], nums2 = [1,3,4,2].
Output: [-1,3,-1]
Explanation:
    For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1.
    For number 1 in the first array, the next greater number for it in the second array is 3.
    For number 2 in the first array, there is no next greater number for it in the second array, so output -1.
Example 2:
Input: nums1 = [2,4], nums2 = [1,2,3,4].
Output: [3,-1]
Explanation:
    For number 2 in the first array, the next greater number for it in the second array is 3.
    For number 4 in the first array, there is no next greater number for it in the second array, so output -1.
Note:
  1. All elements in nums1 and nums2 are unique.
  2. The length of both nums1 and nums2 would not exceed 1000.
X. Ordered stack
这里是建立每个数字和其右边第一个较大数之间的映射,没有的话就是-1。我们遍历原数组中的所有数字,如果此时栈不为空,且栈顶元素小于当前数字,说明当前数字就是栈顶元素的右边第一个较大数,那么建立二者的映射,并且去除当前栈顶元素,最后将当前遍历到的数字压入栈。当所有数字都建立了映射,那么最后我们可以直接通过哈希表快速的找到子集合中数字的右边较大值
https://discuss.leetcode.com/topic/77916/java-10-lines-linear-time-complexity-o-n-with-explanation
Suppose we have a decreasing sequence followed by a greater number
For example [5, 4, 3, 2, 1, 6] then the greater number 6 is the next greater element for all previous numbers in the sequence
We use a stack to keep a decreasing sub-sequence, whenever we see a number x greater than stack.peek() we pop all elements less than x and for all the popped ones, their next greater element is x
For example [9, 8, 7, 3, 2, 1, 6]
The stack will first contain [9, 8, 7, 3, 2, 1] and then we see 6 which is greater than 1 so we pop 1 2 3 whose next greater element should be 6
    public int[] nextGreaterElement(int[] findNums, int[] nums) {
        Map<Integer, Integer> map = new HashMap<>(); // map from x to next greater element of x
        Stack<Integer> stack = new Stack<>();
        for (int num : nums) {
            while (!stack.isEmpty() && stack.peek() < num)
                map.put(stack.pop(), num);
            stack.push(num);
        }   
        for (int i = 0; i < findNums.length; i++)
            findNums[i] = map.getOrDefault(findNums[i], -1);
        return findNums;
    }
https://discuss.leetcode.com/topic/77880/simple-o-m-n-java-solution-using-stack
    public int[] nextGreaterElement(int[] findNums, int[] nums) {
        int[] ret = new int[findNums.length];
        ArrayDeque<Integer> stack = new ArrayDeque<>();
        HashMap<Integer, Integer> map = new HashMap<>();
        for(int i = nums.length - 1; i >= 0; i--) {
            while(!stack.isEmpty() && stack.peek() <= nums[i]) {
                stack.pop();
            }
            if(stack.isEmpty()) map.put(nums[i], -1);
            else map.put(nums[i], stack.peek());
            stack.push(nums[i]);
        }
        for(int i = 0; i < findNums.length; i++) {
            ret[i] = map.get(findNums[i]);
        }
        return ret;
    }

http://bookshadow.com/weblog/2017/02/05/leetcode-next-greater-element-i/
栈stack维护nums的递减子集,记nums的当前元素为n,栈顶元素为top

重复弹出栈顶,直到stack为空,或者top大于n为止

将所有被弹出元素的next greater element置为n
X.
时间复杂度O(n * m) 其中n为nums的长度,m为findNums的长度
https://discuss.leetcode.com/topic/77868/my-concise-short-solution
public int[] nextGreaterElement(int[] findNums, int[] nums) {
 int n1 = findNums.length, n2 = nums.length;
 List<Integer> list = new ArrayList<>();
 for (int i : nums) list.add(i);
 int[] res = new int[n1];
 for (int i = 0; i < n1; i++) {
  int cur = findNums[i];
  res[i] = -1;
  for (int k = list.indexOf(cur); k < n2; k++) {
   if (nums[k] > cur){
    res[i] = nums[k];
    break;
   }
  }
 }
 return res;
}
X. Brute Force
http://blog.csdn.net/u014688145/article/details/70254955

我一开始想都没想,针对nums1的每个元素,循环遍历找nums2中的next greater。很明显它的复杂度就是O(nm)
public int[] nextGreaterElement(int[] findNums, int[] nums) { int[] ans = new int[findNums.length]; for (int i = 0; i < findNums.length; i++) { ans[i] = -1; boolean canFind = false; for (int key : nums) { if (key == findNums[i]) { canFind = true; } if (canFind && key > findNums[i]) { ans[i] = key; break; } } } return ans; }

http://www.cnblogs.com/grandyang/p/6399855.html
    vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {
        vector<int> res(findNums.size());
        for (int i = 0; i < findNums.size(); ++i) {
            int j = 0, k = 0;
            for (; j < nums.size(); ++j) {
                if (nums[j] == findNums[i]) break;
            }
            for (k = j + 1; k < nums.size(); ++k) {
                if (nums[k] > nums[j]) {
                    res[i] = nums[k];
                    break;
                }
            }
            if (k == nums.size()) res[i] = -1;
        }
        return res;
    }

X.
我们来对上面的方法稍做优化,我们用哈希表先来建立每个数字和其坐标位置之间的映射,那么我们在遍历子集合中的数字时,就能直接定位到该数字在原数组中的位置,然后再往右边遍历寻找较大数即可

https://discuss.leetcode.com/topic/77904/easy-to-understand-o-mn-java-solution
public int[] nextGreaterElement(int[] findNums, int[] nums) {
    if(findNums == null ||  nums == null || 
       findNums.length == 0 || nums.length == 0 || 
       findNums.length > nums.length) return new int[0];
    
    int m = findNums.length;
    int n = nums.length;
    int[] result = new int[m];
    Map<Integer, Integer> map = new HashMap<Integer, Integer>();
    
    for(int j = 0; j < n; ++j){
        map.put(nums[j], j);
    }
    for(int i = 0; i < m; ++i){
        int j = map.get(findNums[i]);
        for(; j < n; ++j){
            if(nums[j] > findNums[i]) break;
        }
        result[i] = j < n ? nums[j] : -1;
    }
    return result;
}




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