*Leetcode 81. Search in Rotated Sorted Array II

常规容易想到的解法：

基本就是两点：1、判断在两个上升的段的哪个段 2、mid跟r相等的时候r--可以解决问题的

class Solution {
public:
bool search(vector<int>& nums, int target) {
if(nums.size() == 0) return false;

int l = 0, r = nums.size()-1;
//int end = nums.size() - 1;
while (l <= r) {
int mid = (l + r)/2;

if (target == nums[r] || target == nums[mid]) return true;

if (nums[mid] == nums[r]) {
// for (int j = l; j <= r; j ++) {
// if (nums[j] == target) return true;
// }
// return false;
r --;
continue;
}

if (nums[mid] > nums[r]) {
if (target < nums[mid]) {
if (target > nums[r]) {
r = mid - 1;
} else {
l = mid + 1;
}
} else {

l = mid + 1;
}
} else {
if (target < nums[mid]) {
r = mid - 1 ;
} else {
// target > nums[mid]
if (target > nums[r]) {
r = mid - 1;
} else {
l = mid + 1;
}
}
}
}
return false;
}

};

看到的最好的解法:

class Solution {
public:
bool search(vector<int>& nums, int target) {
int left = 0, right = nums.size()-1, mid;

while(left<=right)
{
mid = (left + right) >> 1;
if(nums[mid] == target) return true;

// the only difference from the first one, trickly case, just updat left and right
if( (nums[left] == nums[mid]) && (nums[right] == nums[mid]) ) {++left; --right;}

else if(nums[left] <= nums[mid])
{
if( (nums[left]<=target) && (nums[mid] > target) ) right = mid-1;
else left = mid + 1;
}
else
{
if((nums[mid] < target) && (nums[right] >= target) ) left = mid+1;
else right = mid-1;
}
}
return false;
}
};