二分查找【Lecode_HOT100】

1.搜索插入位置No.35

class Solution {
    public int searchInsert(int[] nums, int target) {
        int l = 0;
        int r = nums.length-1;
        while(l<=r){
            int mid = (l+r)>>1;
            if(nums[mid]==target){
                return mid;
            }else if(nums[mid]<target){
                l = mid+1;
            }else{
                r = mid-1;
            }
        }
        return l;
    }
}

2.搜索二维矩阵No.74

• 方法一：遍历

public boolean searchMatrix(int[][] matrix, int target) {
        int m = matrix.length;
        int n = matrix[0].length;
        for(int i = 0;i < m;i++){
            for (int j = 0;j < n;j++){
                if(matrix[i][j]==target){
                    return true;
                }
            }
        }
        return false;

    }

• 方法二：二分

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        int m = matrix.length;
        int n = matrix[0].length;

        // 定义二分查找的左右边界
        int left = 0;
        int right = m * n - 1;

        while (left <= right) {
            // 计算中间位置
            int mid = left + (right - left) / 2;
            // 将一维索引转换为二维矩阵中的行和列
            int row = mid / n;
            int col = mid % n;

            int midVal = matrix[row][col];
            if (midVal == target) {
                return true; // 找到目标值
            } else if (midVal < target) {
                left = mid + 1; // 在右半部分查找
            } else {
                right = mid - 1; // 在左半部分查找
            }
        }

        return false; // 未找到目标值
    }
}

3.在排序数组中查找元素的第一个和最后一个位置No.34

• 思路：acwing整数二分

 public int[] searchRange(int[] nums, int target) {
        // 创建返回值数组
        int[] res = new int[2];
        res[0] = -1; // 默认值是 -1
        res[1] = -1; // 默认值是 -1
        int l = 0;
        int r = nums.length - 1;

        // 寻找左边界
        while (l <= r) {
            int mid = (l + r + 1) >> 1;
            if (nums[mid] == target) {
                res[0] = mid;
                r = mid-1;  //继续向左搜索
            } else if (nums[mid] < target) {
                l = mid + 1;
            } else {
                r = mid - 1;
            }
        }
        // 寻找右边界
        l = 0;
        r = nums.length - 1;
        while (l <= r) {
            int mid = (l + r) >> 1;
            if (nums[mid] == target) {
                res[1] = mid;
                l = mid+1; //继续向右搜索
                
            } else if (nums[mid] > target) {
                r = mid - 1;
            } else {
                l = mid + 1;
            }
        }
        return res;
    }

public int[] searchRange(int[] nums, int target) {

        int l = 0;
        int r = nums.length - 1;
        int[] res = new int[2];
        res[0] = -1;
        res[1] = -1;
        if (nums.length == 0 || nums == null)
            return res;

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

        if (nums[l] != target) {
            return res;
        } else {
            res[0] = l;
        }

        l = 0;
        r = nums.length - 1;
        while (l < r) {
            int mid = (l + r + 1) >> 1;//+1代表确认右边界
            if (nums[mid] <= target) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }
        res[1] = l;
        return res;

    }

4.搜索旋转排序数组No.33

• 方法一：for循环遍历

public int search(int[] nums, int target) {
        //for循环遍历
        for(int i = 0;i<nums.length;i++){
            if(nums[i]==target){
                return i;
            }
        }
        return -1;
    }

• 方法二：二分 O（logN）

public int search(int[] nums, int target) {
        int l = 0;
        int r = nums.length-1;
        while(l<=r){
            int mid = (l+r)>>1;
            if(nums[mid]==target){
                return mid;
            }
            if(nums[l]<=nums[mid]){
                //左半边有序
                if(nums[l]<=target&&target<=nums[mid]){
                    r = mid-1;
                }else{
                    l = mid +1;
                }
            }else{
                //右半边有序
                if(nums[mid]<=target&&target<=nums[r]){
                    l = mid+1;
                }else{
                    r = mid - 1;
                }
            }
        }
        return -1;
    }

5.寻找旋转排序数组中的最小值No.153

• 二分
  • 关键在于判断中间值是否大于最右边的值

public int findMin(int[] nums) {
        int l = 0;
        int r = nums.length-1;
        while(l<r){
            int mid = (l+r)>>1;
            if(nums[mid]>=nums[r]){
                //说明小数在右半部分
                l = mid+1;
            }else{
                r = mid;
            }
        }
        return nums[r];
    }