5454
5555<!-- 这里可写通用的实现逻辑 -->
5656
57- 二分查找。
57+ ** 方法一:二分查找**
58+ 59+ 我们使用二分,将数组分割成 ` [left, mid] ` , ` [mid + 1, right] ` 两部分,这时候可以发现,其中有一部分一定是有序的。
60+ 61+ 因此,我们可以根据有序的那一部分,判断 ` target ` 是否在这一部分中:
62+ 63+ - 若 ` [0, mid] ` 范围内的元素构成有序数组:
64+ - 若满足 ` nums[0] <= target <= nums[mid] ` ,那么我们搜索范围可以缩小为 ` [left, mid] ` ;
65+ - 否则,在 ` [mid + 1, right] ` 中查找;
66+ - 若 ` [mid + 1, n - 1] ` 范围内的元素构成有序数组:
67+ - 若满足 ` nums[mid] < target <= nums[n - 1] ` ,那么我们搜索范围可以缩小为 ` [mid + 1, right] ` ;
68+ - 否则,在 ` [left, mid] ` 中查找。
69+ 70+ 二分查找终止条件是 ` l >= r ` ,若结束后发现 ` nums[left] ` 与 ` target ` 不等,说明数组中不存在值为 ` target ` 的元素,返回 -1,否则返回下标 left。
5871
5972<!-- tabs:start -->
6073
6578``` python
6679class Solution :
6780 def search (self nums : List[int ], target : int ) -> int :
68- l, r = 0 , len (nums) - 1
69- while l <= r:
70- mid = (l + r) >> 1
71- if nums[mid] == target:
72- return mid
73- if nums[mid] > target:
74- if nums[mid] >= nums[r] and target < nums[l]:
75- l = mid + 1
81+ n = len (nums)
82+ left, right = 0 , n - 1
83+ while left < right:
84+ mid = (left + right) >> 1
85+ if nums[0 ] <= nums[mid]:
86+ if nums[0 ] <= target and target <= nums[mid]:
87+ right = mid
7688 else :
77- r = mid - 1
89+ left = mid + 1
7890 else :
79- if nums[mid] <= nums[l] and target > nums[r ]:
80- r = mid - 1
91+ if nums[mid] < target and target <= nums[n - 1 ]:
92+ left = mid + 1
8193 else :
82- l = mid+ 1
83- return - 1
94+ right = mid
95+ return left if nums[left] == target else - 1
8496```
8597
8698### ** Java**
@@ -90,19 +102,25 @@ class Solution:
90102``` java
91103class Solution {
92104 public int search (int [] nums , int target ) {
93- int l = 0 , r = nums. length - 1 ;
94- while (l <= r) {
95- int mid = (l + r) >>> 1 ;
96- if (nums[mid] == target) return mid;
97- if (nums[mid] > target) {
98- if (nums[mid] >= nums[r] && target < nums[l]) l = mid + 1 ;
99- else r = mid - 1 ;
105+ int n = nums. length;
106+ int left = 0 , right = n - 1 ;
107+ while (left < right) {
108+ int mid = (left + right) >> 1 ;
109+ if (nums[0 ] <= nums[mid]) {
110+ if (nums[0 ] <= target && target <= nums[mid]) {
111+ right = mid;
112+ } else {
113+ left = mid + 1 ;
114+ }
100115 } else {
101- if (nums[mid] <= nums[l] && target > nums[r]) r = mid - 1 ;
102- else l = mid + 1 ;
116+ if (nums[mid] < target && target <= nums[n - 1 ]) {
117+ left = mid + 1 ;
118+ } else {
119+ right = mid;
120+ }
103121 }
104122 }
105- return - 1 ;
123+ return nums[left] == target ? left : - 1 ;
106124 }
107125}
108126```
@@ -115,23 +133,56 @@ class Solution {
115133class Solution {
116134public:
117135 int search(vector<int >& nums, int target) {
118- int l = 0, r = nums.size() - 1;
119- while (l <= r) {
120- int mid = (l + r) >> 1;
121- if (nums[ mid] == target) return mid;
122- if (nums[ mid] > target) {
123- if (nums[ mid] >= nums[ r] && target < nums[ l] ) l = mid + 1;
124- else r = mid - 1;
125- } else {
126- if (nums[ mid] <= nums[ l] && target > nums[ r] ) r = mid - 1;
127- else l = mid + 1;
136+ int n = nums.size();
137+ int left = 0, right = n - 1;
138+ while (left < right)
139+ {
140+ int mid = (left + right) >> 1;
141+ if (nums[ 0] <= nums[ mid] )
142+ {
143+ if (nums[ 0] <= target && target <= nums[ mid] ) right = mid;
144+ else left = mid + 1;
145+ }
146+ else
147+ {
148+ if (nums[ mid] < target && target <= nums[ n - 1] ) left = mid + 1;
149+ else right = mid;
128150 }
129151 }
130- return -1;
152+ return nums [ left ] == target ? left : -1;
131153 }
132154};
133155```
134156
157+ ### **Go**
158+
159+ ```go
160+ func search(nums []int, target int) int {
161+ n := len(nums)
162+ left, right := 0, n-1
163+ for left < right {
164+ mid := (left + right) >> 1
165+ if nums[0] <= nums[mid] {
166+ if nums[0] <= target && target <= nums[mid] {
167+ right = mid
168+ } else {
169+ left = mid + 1
170+ }
171+ } else {
172+ if nums[mid] < target && target <= nums[n-1] {
173+ left = mid + 1
174+ } else {
175+ right = mid
176+ }
177+ }
178+ }
179+ if nums[left] == target {
180+ return left
181+ }
182+ return -1
183+ }
184+ ```
185+ 135186### ** JavaScript**
136187
137188``` js
@@ -141,29 +192,26 @@ public:
141192 * @return {number}
142193 */
143194var search = function (nums , target ) {
144- let l = 0,
145- r = nums.length - 1;
146- if (l > r) return -1;
147- while (l <= r) {
148- let mid = l + Math.floor((r - l) / 2);
149- if (nums[mid] === target) return mid;
150- else if (
151- nums[mid] <= nums[r] &&
152- target <= nums[r] &&
153- target >= nums[mid]
154- )
155- l = mid + 1;
156- else if (
157- nums[mid] >= nums[l] &&
158- target <= nums[mid] &&
159- target >= nums[l]
160- )
161- r = mid - 1;
162- else if (nums[mid] >= nums[r]) l = mid + 1;
163- else if (nums[mid] <= nums[l]) r = mid - 1;
164- else return -1;
195+ const n = nums .length ;
196+ let left = 0 ,
197+ right = n - 1 ;
198+ while (left < right) {
199+ const mid = (left + right) >> 1 ;
200+ if (nums[0 ] <= nums[mid]) {
201+ if (nums[0 ] <= target && target <= nums[mid]) {
202+ right = mid;
203+ } else {
204+ left = mid + 1 ;
205+ }
206+ } else {
207+ if (nums[mid] < target && target <= nums[n - 1 ]) {
208+ left = mid + 1 ;
209+ } else {
210+ right = mid;
211+ }
212+ }
165213 }
166- return -1;
214+ return nums[left] == target ? left : - 1 ;
167215};
168216```
169217
@@ -199,6 +247,33 @@ impl Solution {
199247}
200248```
201249
250+ ### ** TypeScript**
251+ 252+ ``` ts
253+ function search(nums : number [], target : number ): number {
254+ const = nums .length ;
255+ let left = 0 ,
256+ right = n - 1 ;
257+ while (left < right ) {
258+ const = (left + right ) >> 1 ;
259+ if (nums [0 ] <= nums [mid ]) {
260+ if (nums [0 ] <= target && target <= nums [mid ]) {
261+ right = mid ;
262+ } else {
263+ left = mid + 1 ;
264+ }
265+ } else {
266+ if (nums [mid ] < target && target <= nums [n - 1 ]) {
267+ left = mid + 1 ;
268+ } else {
269+ right = mid ;
270+ }
271+ }
272+ }
273+ return nums [left ] == target ? left : - 1 ;
274+ }
275+ ```
276+ 202277### ** ...**
203278
204279```
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