二分查找题总结

hot100

hot100

搜索插入位置

代码：

java 复制代码

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

    }
}

搜索二维矩阵

题目链接：
74.搜索二维矩阵

代码：

java 复制代码

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {

        int m = matrix.length, n = matrix[0].length;
        int l = 0, r = m*n - 1;
        while (l <= r)
        {
            int mid = l + (r-l)/2;
            int row = mid / n;
            int col = mid % n;
            if (target == matrix[row][col]) return true;
            if (target > matrix[row][col]) l = mid + 1;
            if (target < matrix[row][col]) r = mid - 1;
        }
        return false;

    }
}

在排序数组中查找元素的第一个和最后一个位置

题目链接：
34.在排序数组中查找元素的第一个和最后一个位置

代码：

java 复制代码

class Solution {
    int binarySearch(int[] nums, int target){
        int left = 0, right = nums.length - 1;
        while (left <= right){
            int mid = left + (right - left) / 2;
            if (nums[mid] == target){
                return mid;
            }else if (nums[mid] > target){
                right = mid - 1;
            }else if (nums[mid] < target){
                left = mid + 1;
            }
        }
        return -1;
    }
    public int[] searchRange(int[] nums, int target) {
        int index = binarySearch(nums,target);
        if (index == -1){
            return new int[]{-1,-1};
        }
        int left = index;
        int right = index;

        while(left - 1 >= 0 && nums[left] == nums[left - 1]){
            left --;
        }
        while(right + 1 < nums.length && nums[right] == nums[right + 1]){
            right ++;
        }
        return new int[]{left,right};
    }
}

搜索旋转排序数组

题目链接：
33.搜索旋转排序数组

代码：

java 复制代码

class Solution {
    public int search(int[] nums, int target) {
        int n = nums.length;
        if (n == 0) return -1;
        if (n == 1) return nums[0] == target ? 0:-1;
        
        int l = 0, r = n - 1;
        while (l <= r)
        {
            int mid = l + (r - l) / 2;
            if (target == nums[mid]) return mid;
            if (nums[0] <= nums[mid])
            {
                if (nums[0] <= target && target < nums[mid])
                {
                    r = mid - 1;
                }
                else
                {
                    l = mid + 1;
                }
            }
            else
            {
                if (nums[mid] < target && target <= nums[n - 1])
                {
                    l = mid + 1;
                }
                else
                {
                    r = mid - 1;
                }
            }
        }
        return -1;

    }
}

寻找旋转排序数组中的最小值

题目链接：
153.寻找旋转排序数组中的最小值

代码：

java 复制代码

class Solution {
    public int findMin(int[] nums) {
        int l = 0, r = nums.length - 1;
        int minn = Integer.MAX_VALUE;

        while (l <= r) {
            int mid = l + (r - l) / 2;
            if (nums[mid] < nums[r]) {
                minn = Math.min(minn, nums[mid]);
                r = mid - 1;
            }else {
                minn = Math.min(minn, nums[l]);
                l = mid + 1;
            }
        }
        return minn;
    }
}

寻找两个正序数组的中位数

题目链接：
4.寻找两个正序数组的中位数

代码：

java 复制代码

class Solution {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        int m = nums1.length, n = nums2.length;
        int len = m + n;
        int left = -1, right = -1;
        int aStart = 0, bStart = 0;
        
        for (int i = 0; i <= len / 2; i ++)
        {
            left = right;
            if (aStart < m && (bStart >= n || nums1[aStart] < nums2[bStart]))
            {
                right = nums1[aStart ++];
            }
            else
            {
                right = nums2[bStart ++];
            }
        }
        if (len % 2 == 0)
        {
            return (left + right) / 2.0;
        }
        else
        {
            return right;
        }

    }
}