503. Next Greater Element II
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Given a circular array (the next element of the last element is the first element of the array), print the Next Greater Number for every element. The Next Greater Number of a number x is the first greater number to its traversing-order next in the array, which means you could search circularly to find its next greater number. If it doesn’t exist, output -1 for this number.
Example 1:
Input: [1,2,1]
Output: [2,-1,2]
Explanation: The first 1’s next greater number is 2;
The number 2 can’t find next greater number;
The second 1’s next greater number needs to search circularly, which is also 2.
Note: The length of given array won’t exceed 10000.
题解
环形数组,找出每个元素之后第一个比它大的数
解法1
下面是我最开始的思路:
如果从前往后遍历,对每个数都往后找寻第一个比它大的数,时间复杂度O(n^2),会超时。
为了避免重复工作,我们从后往前遍历,这样前面的数可以利用后面的数的结果。比如nums[i]的第一个比它大的数的下标为k,那么对nums[i-1],如果nums[i-1] < nums[i],那么结果的下标就是i,等于则结果的下标就是k,大于则结果的下标最小是k,直接从k开始找。
代码如下,用时22ms
public class Solution { public int[] nextGreaterElements(int[] nums) { int n = nums.length, temp[] = new int[n];//temp记录第一个大的数的下标 for(int i = 0; i < n; i++) temp[i] = -2;//-2代表未知 //从后往前找,尽量避免重复工作 for(int i = n - 1; i >= 0; i--){ int j = i + 1; while(j < i + n){ //比nums[i]大,找到 if(nums[j % n] > nums[i]) break; else{ //-1说明已经没有比num[j]更大的数,直接跳过所有数 if(temp[j % n] == -1) j = i + n; //-2说明未知,继续下一个数 else if(temp[j % n] == -2) j++; //直接跳到比nums[j]大的第一个数,再与nums[i]比较 else j = temp[j % n]; } } temp[i] = (j == i + n ? -1 : j); } //得到结果 int[] res = new int[n]; for(int i = 0; i < n; i++){ res[i] = (temp[i] == -1 ? -1 : nums[temp[i] % n]); } return res; }}
解法2
维持一个从栈顶向下从小到大排列的栈来实现,从前向后遍历,每次把所有比当前数num[i]小的数出栈,并设置它们的第一个大的数的下标为i,之后把当前数入栈。
代码如下,用时59 ms
public class Solution { public int[] nextGreaterElements(int[] nums) { int n = nums.length, next[] = new int[n]; Arrays.fill(next, -1); Stack<Integer> stack = new Stack<>(); // index stack for (int i = 0; i < n * 2; i++) { int num = nums[i % n]; while (!stack.isEmpty() && nums[stack.peek()] < num) next[stack.pop()] = num; if (i < n) stack.push(i); } return next; }}
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