solve #154

Author: aylei

src/lib.rs

@@ -148,3 +148,4 @@ mod n0150_evaluate_reverse_polish_notation;
 mod n0151_reverse_words_in_a_string;
 mod n0152_maximum_product_subarray;
 mod n0153_find_minimum_in_rotated_sorted_array;
+mod n0154_find_minimum_in_rotated_sorted_array_ii;

src/n0154_find_minimum_in_rotated_sorted_array_ii.rs

@@ -0,0 +1,76 @@
+/**
+ * [154] Find Minimum in Rotated Sorted Array II
+ *
+ * Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
+ * 
+ * (i.e.,  [0,1,2,4,5,6,7] might become  [4,5,6,7,0,1,2]).
+ * 
+ * Find the minimum element.
+ * 
+ * The array may contain duplicates.
+ * 
+ * Example 1:
+ * 
+ * 
+ * Input: [1,3,5]
+ * Output: 1
+ * 
+ * Example 2:
+ * 
+ * 
+ * Input: [2,2,2,0,1]
+ * Output: 0
+ * 
+ * Note:
+ * 
+ * 
+ * 	This is a follow up problem to <a href="https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/description/">Find Minimum in Rotated Sorted Array</a>.
+ * 	Would allow duplicates affect the run-time complexity? How and why?
+ * 
+ * 
+ */
+pub struct Solution {}
+
+// submission codes start here
+
+/*
+ 针对无重复的做法, 只要二分搜索找折点即可: 假如 nums[mid] > nums[base] 那么转折点一定在右侧, 否则在左侧
+
+ 但假如有重复, 就可能有 nums[mid] == nums[base], 这时就尴尬了, 无法确定转折点在左半部分还是右半部分
+
+ 可以考虑一个数组, [1,1,1,1,1,1,1,0,1,1,1,1,1,1] 这个数组无论怎么去找 0, 时间复杂度无法低于 O(N)
+
+ 但假如不是这种极端情况, 那么二分搜索还是能优化的, 在 153 的基础上, 碰到相等就跳过即可
+ */
+impl Solution {
+    pub fn find_min(nums: Vec<i32>) -> i32 {
+        let mut lo = 0;
+        let mut hi = nums.len() - 1;
+        let mut mid = 0;
+        while lo < hi {
+            mid = lo + (hi - lo) / 2;
+            if (nums[mid] > nums[hi]) {
+                lo = mid + 1;
+            } else if (nums[mid] < nums[hi]) {
+                hi = mid;
+            } else {
+                hi -= 1;
+            }
+        }
+        return nums[lo]
+    }
+}
+
+// submission codes end
+
+#[cfg(test)]
+mod tests {
+    use super::*;
+
+    #[test]
+    fn test_154() {
+        assert_eq!(Solution::find_min(vec![1,2,2,2,2,2]), 1);
+        assert_eq!(Solution::find_min(vec![1,3,3]), 1);
+        assert_eq!(Solution::find_min(vec![3,1,3,3]), 1);
+    }
+}