算法必刷系列之查找、排序

二分查找

顺序查找

逐个遍历，判断是否是待查找的元素

public int search(int[] array,int key){
    for(int i=0;i<array.length;i++){
        if(array[i]==key){
            return i;
        }
    }
    return -1;
}
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二分查找迭代写法

有序数组可用，可作为模板记忆，注意边界

public int binarySearch(int[] array,int key,int low,int high){
    while (low<= high){
        int mid = low+((high-low)>>1);
        if(array[mid]==key){
            return mid;
        }else if(array[mid]>key){
            high = mid - 1;
        }else if(array[mid]<key){
            low = mid + 1;
        }
    }
    return -1;
}
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二分查找递归写法

有序数组可用，可作为模板记忆，注意边界

public int binarySearch(int[] array,int key,int low,int high){
    if(low<=high){
        int mid = low+((high-low)>>1);
        if(array[mid]==key){
            return mid;
        }else if(array[mid]>key){
            return binarySearch(array,key,low,mid-1);
        }else if(array[mid]<key){
            return binarySearch(array,key,mid+1,high);
        }
    }
    return -1;
}
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元素中有重复元素的二分查找

基于模板找到元素后，继续向前遍历，找到等于待查找元素在数组中首次出现的位置，如果元素过多，也可使用二分查找方式继续查找

public int binarySearchRe(int[] array, int key, int low, int high) {
    while (low <= high) {
        int mid = low + ((high - low) >> 1);
        if (array[mid] == key) {
            while (mid >= 0 && array[mid] == key) {
                mid--;
            }
            return mid + 1;
        } else if (array[mid] > key) {
            high = mid - 1;
        } else if (array[mid] < key) {
            low = mid + 1;
        }
    }
    return -1;
}
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在排序数组中查找元素的第一个和最后一个位置

基于模板找到元素后，继续向前、后遍历，找到等于待查找元素在数组中首次出现的位置和最后出现的位置，如果元素过多，也可使用二分查找方式继续查找

public int[] searchRange(int[] nums, int target) {
    return binarySearch(nums,target,0,nums.length-1);
}

public int[] binarySearch(int[] nums,int target,int low,int high){
    int[] res = new int[]{-1,-1};
    while(low<=high){
        int mid = low + ((high-low)>>1);
        if(nums[mid]==target){
            int left = mid;
            while(left>=0&&nums[left]==target){
                left--;
            }
            int right = mid;
            while(right<nums.length&&nums[right]==target){
                right++;
            }
            res[0] = left+1;
            res[1] = right-1;
            return res;
        }else if(nums[mid]>target){
            high = mid-1;
        }else if(nums[mid]<target){
            low = mid+1;
        }
    }
    return res;
}
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山脉数组的峰顶索引

注意山脉的判断条件，高于两侧数据，递增顺序在左侧，递减顺序在右侧

public int peakIndexInMountainArray(int[] arr) {
    int high = arr.length-2;
    int low = 1;
    while(low <= high){
        int mid = low+((high-low)>>1);
        if(arr[mid]>arr[mid-1] && arr[mid]>arr[mid+1]){
            return mid;
        }else if(arr[mid]>arr[mid-1] && arr[mid]<arr[mid+1]){
            low = mid+1;
        }else if(arr[mid]<arr[mid-1] && arr[mid]>arr[mid+1]){
            high = mid - 1;
        }
    }
    return -1;
}
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旋转数字的最小数字

使用二分查找模板，边界值通过测试样例确定

public int findMin(int[] nums) {
    int low = 0;
    int high = nums.length - 1;
    while(low<high){
        int mid = low+((high-low)>>1);
        if(nums[mid]>nums[high]){
            low = mid+1;
        }else if(nums[mid]<nums[high]){
            high = mid;
        }
    }
    return nums[low];
}
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找缺失数字

使用二分查找模板，边界值通过测试样例确定

public int missingNumber(int[] array){
    int high = array.length;
    int low = 0;
    while (low<=high){
        int mid = low + ((high - low)>>1);
        if(array[mid] == mid){
            low = mid+1;
        }else{
            high = mid - 1;
        }
    }
    return low;
}
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优化求平方根

使用二分查找模板，边界值通过测试样例确定

public int mySqrt(int x) {
    int low = 1;
    int high = x;
    while(low <= high){
        int num = low + ((high-low)>>1);
        if(x/num == num){
            return num;
        }else if(x/num>num){
            low = num + 1;
        }else{
            high = num - 1;
        }
    }
    return high;
}
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二叉搜索树中搜索指定值

二分查找与二叉树的结合，先判断根节点，根据根节点与待查找值的情况判断在左子树或者右子树查找

public TreeNode searchBST(TreeNode root, int val) {
    if(root==null){
        return null;
    }else if(root.val == val){
        return root;
    }else if(root.val<val){
        return searchBST(root.right,val);
    }else if(root.val>val){
        return searchBST(root.left,val);
    }
    return null;
}
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验证二叉搜索树

限定low和high两个范围进行验证，验证左右子树时，根据根节点的值更新low和high

public boolean isValidBST(TreeNode root) {
    long low = Long.MIN_VALUE;
    long high = Long.MAX_VALUE;
    return isValidBST(root,low,high);
}

public boolean isValidBST(TreeNode root,long low,long high){
    if(root==null){
        return true;
    }
    if(root.val<=low || root.val >=high){
        return false;
    }
    return isValidBST(root.left,low,root.val)&&isValidBST(root.right,root.val,high);
}
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有序数组转化为二叉搜索树

每次选择中间值作为根节点，再通过左右两个序列创建左右子树

public TreeNode sortedArrayToBST(int[] nums) {
    return sortedArrayToBST(nums,0,nums.length-1);
}

public TreeNode sortedArrayToBST(int[] nums,int left,int right){
    if(left>right){
        return null;
    }
    int mid = left+((right-left)>>1);
    TreeNode root = new TreeNode(nums[mid]);
    root.left = sortedArrayToBST(nums,left,mid-1);
    root.right = sortedArrayToBST(nums,mid+1,right);
    return root;
}
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快速排序

以第一个元素为基准实现快速排序

快速排序模板1

public void quickSort(int[] array, int left, int right) {
    if(left<right){
        int pivot = array[left];
        int i = right + 1;
        for (int j = right; j > left; j++) {
            if (array[j] > pivot) {
                i--;
                int temp = array[j];
                array[j] = array[i];
                array[i] = temp;
            }
        }
        int pivotIndex = i - 1;
        int temp = array[left];
        array[left] = array[pivotIndex];
        array[pivotIndex] = temp;
        quickSort(array,left,pivotIndex-1);
        quickSort(array,pivotIndex+1,right);
    }
}
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以最后一个元素为基准实现快速排序

快速排序模板2

public void quickSort(int[] array, int left, int right) {
    if (left < right) {
        int pivot = array[right];
        int i = left - 1;
        for (int j = left; j < right; j++) {
            if (array[j] < pivot) {
                i++;
                int temp = array[i];
                array[i] = array[j];
                array[j] = temp;
            }
        }
        int pivotIndex = i + 1;
        int temp = array[pivotIndex];
        array[pivotIndex] = array[right];
        array[right] = temp;
        quickSort(array, left, pivotIndex - 1);
        quickSort(array, pivotIndex + 1, right);
    }
}
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以中间元素为基准实现快速排序

快速排序模板3

public static void quickSort(int[] array, int start, int end) {
    if (start >= end) {
        return;
    }
    int left = start;
    int right = end;
    int mid = (left + right) / 2;
    int pivot = array[mid];
    while (left <= right) {
        while (left <= right && array[left] < pivot) {
            left++;
        }
        while (left <= right && array[right] > pivot) {
            right--;
        }
        if (left <= right) {
            int temp = array[left];
            array[left] = array[right];
            array[right] = temp;
            left++;
            right--;
        }
    }
    quickSort(array, start, right);
    quickSort(array, left, end);
}
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归并排序

归并排序

归并排序模板

public void mergeSort(int[] array, int start, int end, int[] temp) {
	//2.直至每个序列只包含一个元素，停止划分
    if (start >= end) {
        return;
    }
    //1.从中间开始，每次划分为两个序列
    mergeSort(array, start, (start + end) / 2, temp);
    mergeSort(array, (start + end) / 2 + 1, end, temp);
    //3。进行有序数组的合并
    merge(array, start, end, temp);
}

public void merge(int[] array, int start, int end, int[] temp) {
	//找到序列中点
    int mid = (start + end) / 2;
    //left遍历左边的序列
    int left = start;
    //right遍历右边的序列
    int right = mid + 1;
    //index遍历临时数组，存储合并结果
    int index = 0;
    //两个序列均从起点到终点进行遍历
    while (left <= mid && right <= end) {
    	//将两个序列中较小的元素放入临时数组中
        if (array[left] < array[right]) {
            temp[index++] =array[left++];
        }else {
            temp[index++] =array[right++];
        }
    }
    //此时仅剩一个序列未遍历结束，直接赋值
    while (left <= mid){
        temp[index++] =array[left++];
    }
    while (right<=end){
        temp[index++] =array[right++];
    }
    //将归并的结果拷贝到原数组
    for (int i=start;i<=end;i++){
        array[i] = temp[i];
    }
}
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数组中第K大的数的问题可以通过数组排序解决，根据题目中给定的时间、空间复杂福选择合适的排序算法