【优选算法】专题三:二分查找(二)

69. x 的平方根

69. x 的平方根

class Solution {
    public int mySqrt(int x) {
        // 注意细节处理0的问题
        if(x < 1) return 0;
        long left = 0,right = x;
        while(left < right){
            long mid = left + (right -left+1)/2;
            if(mid * mid <=x) left = mid ;
            else right = mid -1;
        }
        return (int)left;
    }
}
1
2
3
4
5
6
7
8
9
10
11
12
13

35. 搜索插入位置

35. 搜索插入位置

class Solution {
    public int searchInsert(int[] nums, int target) {
        int left = 0,right = nums.length-1;
        while(left<right){
            int mid = left +(right-left)/2;
            if(nums[mid] < target) left = mid+1;
            else right =mid;
        }
        //处理第三种情况 
        if(nums[left]<target) return left+1;
        return left;
    }
}
1
2
3
4
5
6
7
8
9
10
11
12
13

852. 山脉数组的峰顶索引

852. 山脉数组的峰顶索引

class Solution {
    public int peakIndexInMountainArray(int[] arr) {
        int left=0,right=arr.length-1;
        while(left<right){
            int mid = left+(right-left+1)/2;
            if(arr[mid]>arr[mid-1])left = mid;
            else right = mid-1;
        }
        return left;
    }
}
1
2
3
4
5
6
7
8
9
10
11

162. 寻找峰值

162. 寻找峰值

class Solution {
    public int findPeakElement(int[] arr) {
        int left=0,right=arr.length-1;
        while(left<right){
            int mid = left+(right-left)/2;
            if(arr[mid]>arr[mid+1]) right = mid;
            else left = mid+1;
        }
        return left;
    }
}
1
2
3
4
5
6
7
8
9
10
11

153. 寻找旋转排序数组中的最小值

153. 寻找旋转排序数组中的最小值

class Solution {
    public int findMin(int[] nums) {
        int left=0,right=nums.length-1;
        int n=nums.length-1; 
        while(left<right){
            int mid=left+(right-left)/2;
            if(nums[mid]>nums[n]) left=mid+1;
            else right=mid;
        }
    return nums[left];
    }
}
1
2
3
4
5
6
7
8
9
10
11
12

我们再讨论一下是否能将A点作为参考点:

class Solution {
    public int findMin(int[] nums) {
        int left=0,right=nums.length-1;
        if(nums[0]<nums[right])return nums[0];
        while(left<right){
            int mid=left+(right-left)/2;
            if(nums[mid]>=nums[0]) left=mid+1;
            else right=mid;
        }     
        return nums[left];
    }
}
1
2
3
4
5
6
7
8
9
10
11
12

LCR 173. 点名

LCR 173. 点名

class Solution {
    public int takeAttendance(int[] nums) {
        int left=0,right=nums.length-1;
        if(nums[right]==right) return right+1;
        while(left<right){
            int mid=left+(right-left)/2;
            if(nums[mid]==mid) left=mid+1;
            else right=mid;
        }
        return left;
    }
}
1
2
3
4
5
6
7
8
9
10
11
12