【LeetCode】581. Shortest Unsorted Continuous Subarray
【LeetCode】581. Shortest Unsorted Continuous Subarray
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本文链接：https://blog.csdn.net/shiliang97/article/details/102372447
Given an integer array, you need to find one continuous subarray that if you only sort this subarray in ascending order, then the whole array will be sorted in ascending order, too.
You need to find the shortest such subarray and output its length.
Example 1: Input: [2, 6, 4, 8, 10, 9, 15] Output: 5 Explanation: You need to sort [6, 4, 8, 10, 9] in ascending order to make the whole array sorted in ascending order. Note: Then length of the input array is in range [1, 10,000]. The input array may contain duplicates, so ascending order here means <=.
来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/shortest-unsorted-continuous-subarray 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
int FindUnsortedSubarray(vector<int>& nums) {
//此解法参考英文官方LeetCode上的讨论
//从左到右扫描（或从右到左）找局部极大值（或局部极小值），若位置放置不正确，找到其应该存在的地方
int n = nums.size();
//赋初始开始和结束值
int start = -1;
int end = -2;
//结束值赋为-2是考虑在数组本身就是有序时,return也可以给出正确值
int min = nums[n - 1];
int max = nums[0];
for(int i = 0, pos = 0; i < n; i++) {
pos = n - 1 - i;
//找出局部极大、极小值
max = Max(max, nums[i]);
min = Min(min, nums[pos]);
//如果当前值小于局部极大值，用end记录该位置，找到该max的合适位置，
if(nums[i] < max)
end = i;
//如果当前值大于局部极小值，用star记录该位置，找到该star的合适位置
if(nums[pos] > min)
start = pos;
}
//返回开始和结束的索引差
return end - start + 1;
}