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docs: add a description of the solution to lcci problem: No.08.03 #760

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14 changes: 14 additions & 0 deletions lcci/08.03.Magic Index/README.md
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<!-- 这里可写通用的实现逻辑 -->

**线性查找：**

遍历数组，当 `A[i] = i` 时直接返回即可。

**优化：**

在遍历的基础，进行可能的 "跳跃"，结束时执行 `i = max(A[i], i + 1)`，而不再单纯 `i++`。

可行性证明：

因为数组是**有序**的，若 `A[i] != i`，那么就可以将 `A[i]` 以下的可能全部排除，直接将 `i` 设定为 `A[i]`。

若是考虑最糟的状况（所有元素都为负数），则该优化与遍历无差别。

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### **Python3**
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