方法一:暴力解法
- 要歸納出:
- 找到每個數左邊第 1 個嚴格大於它的值的下標(構成逆序關係);
- 找到每個數右邊第 1 個嚴格小於它的值的下標(也構成逆序關係)。
二者之差就是題目要找的「最短無序連續子數組」。
Java 代碼:
public class Solution {
public int findUnsortedSubarray(int[] nums) {
int len = nums.length;
if (len < 2) {
return 0;
}
// 從左到右找出最後一個單調不減的下標
// 從右到左找出最後一個單調不增的下標
int max = nums[0];
int right = 0;
for (int i = 1; i < len; i++) {
max = Math.max(max, nums[i]);
if (nums[i] < max) {
right = i;
}
}
// System.out.println(right);
int left = len - 1;
int min = nums[len - 1];
for (int i = len - 2; i >= 0; i--) {
min = Math.min(min, nums[i]);
if (nums[i] > min) {
left = i;
}
}
// System.out.println(left);
if (right > left) {
return right - left + 1;
}
return 0;
}
public static void main(String[] args) {
Solution solution = new Solution();
int[] nums = {2, 6, 4, 8, 10, 9, 15};
int res = solution.findUnsortedSubarray(nums);
System.out.println(res);
}
}
方法二:棧(單調棧)
Java 代碼:
import java.util.ArrayDeque;
import java.util.Deque;
public class Solution {
public int findUnsortedSubarray(int[] nums) {
int len = nums.length;
if (len < 2) {
return 0;
}
int leftBound = len - 1;
int rightBound = 0;
Deque<Integer> stack = new ArrayDeque<>(len);
for (int i = 0; i < len; i++) {
// 這裏的 while 不要忘記
while (!stack.isEmpty() && nums[i] < nums[stack.peekLast()]) {
leftBound = Math.min(leftBound, stack.removeLast());
}
stack.addLast(i);
}
stack.clear();
for (int i = len - 1; i >= 0; i--) {
while (!stack.isEmpty() && nums[i] > nums[stack.peekLast()]) {
rightBound = Math.max(rightBound, stack.removeLast());
}
stack.addLast(i);
}
if (rightBound > leftBound) {
return rightBound - leftBound + 1;
}
return 0;
}
}