[Leetcode 1064] Fixed Point

原题说明

Given an array A of distinct integers sorted in ascending order, return the smallest index i that satisfies A[i] == i.  Return -1 if no such i exists.

 

Example 1:

Input: [-10,-5,0,3,7]
Output: 3
Explanation: 
For the given array, A[0] = -10, A[1] = -5, A[2] = 0, A[3] = 3, thus the output is 3.

Example 2:

Input: [0,2,5,8,17]
Output: 0
Explanation: 
A[0] = 0, thus the output is 0.

Example 3:

Input: [-10,-5,3,4,7,9]
Output: -1
Explanation: 
There is no such i that A[i] = i, thus the output is -1.

 

Note:

1. 1 <= A.length < 10^4
2. -10^9 <= A[i] <= 10^9

解题思路

因为给出的是一个排序的array，而index也是自然从0到n-1的排序数组，因此本题采用二分法来查找A[i] == i的index。
当 i < A[i], 我们往左边查找，当 i >= A[i]时， 我们往右边查找。
需要注意的因为要求最小的index，我们更新右端点时，需要 i >= A[i]。如果是最大的index， 则应该是 i > A[i]。

示例代码 (cpp)

class Solution {
public:
    int fixedPoint(vector<int>& A) {
        int left = 0;
        int right = A.size() - 1;
        while (left + 1 < right) {
            int mid = left + (right - left) / 2;
            if (A[mid] >= mid) {
                right = mid;
            }
            else {
                left = mid;
            }
        }
        if (left == A[left]) {
            return left;
        }
        if (right == A[right]) {
            return right;
        }
        return -1;
    }
};

示例代码 (java)

class Solution {
    public int fixedPoint(int[] A) {
        int l = 0, r = A.length - 1;
        while (l < r) {
            int m = (l + r) / 2;
            if (A[m] - m < 0)
                l = m + 1;
            else
                r = m;
        }
        return A[l] == l ? l : -1;
    }
}

示例代码 (python)

class Solution(object):
    def fixedPoint(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        l, r = 0, len(A) - 1
        while l < r:
            m = (l + r) / 2
            if A[m] - m < 0:
                l = m + 1
            else:
                r = m
        return l if A[l] == l else -1

复杂度分析

时间复杂度: O(logN)
空间复杂度: O(1)

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