第二天| 977.有序数组的平方,209.长度最小的子数组,59.螺旋矩阵II

ThefirstBob

已于 2023-11-15 03:04:25 修改

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文章标签：算法矩阵 java

于 2023-11-15 03:02:31 首次发布

本文链接：https://blog.youkuaiyun.com/ThefirstBob/article/details/134411333

文章介绍了在LeetCode平台上两个问题的解决方案，分别涉及使用双指针技巧对数组中的平方数进行排序以及寻找满足条件的子数组和。通过实例展示了`sortedSquares`和`minSubArrayLen`函数的实现，强调了双指针在优化搜索效率中的应用。

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# 977_题目考察点：双指针

## 题目链接：力扣（LeetCode）官网 - 全球极客挚爱的技术成长平台

### 代码实现

package LeetCode;

public class YangSibo_977 {
    public static void main(String args []) {
        int [] nums = {-4,-1,0,3,10};
        int [] res1 = new int[nums.length];
        int [] res2 = new int[nums.length];
        YangSibo_977_1 demo1 = new YangSibo_977_1();
        YangSibo_977_2 demo2 = new YangSibo_977_2();
        res1 = demo1.sortedSquares(nums);
        res2 = demo1.sortedSquares(nums);
        for (int i = 0; i < nums.length; i++) {
            System.out.print(res1[i]);
            System.out.print(",");
        }
        System.out.println();
        for (int i = 0; i < nums.length; i++) {
            System.out.print(res2[i]);
            System.out.print(",");
        }
    }
}

class YangSibo_977_1 {
    public int[] sortedSquares(int[] nums) {
        int length = nums.length;
        int [] res = new int[length];
        if (length == 0) {
            return res;
        }
        if (length == 1) {
            int temp = nums[0] * nums[0];
            nums[0] = temp;
            return nums;
        }
        int [] squaresdata = new int[length];
        for(int i = 0; i < length; i++) {
            squaresdata[i]= nums[i] * nums[i];
        }
        for(int i = 0; i < length -1 ; i++) {
            for(int j = 0; j < length - i -1; j++) {
                if (squaresdata [j+1] < squaresdata[j]) {
                    int temp = squaresdata[j+1];
                    squaresdata[j+1] = squaresdata[j];
                    squaresdata[j] = temp;
                }
            }
        }
        return squaresdata;
    }
}
class YangSibo_977_2 {
    public int[] sortedSquares(int[] nums) {
        int length = nums.length;
        int [] res = new int[length];
        if (length == 0) {
            return res;
        }
        if (length == 1) {
            int temp = nums[0] * nums[0];
            nums[0] = temp;
            return nums;
        }
        int right = 0;
        int left = length - 1;
        int label = length - 1;
        while (right <= left) {
            int rightval = nums[right] * nums[right];
            int leftval = nums[left] * nums[left];
            if( rightval > leftval ) {
                res[label] = rightval;
                right++;
            }else {
                res[label] = leftval;
                left--;
            }
            label--;
        }
        return res;
    }
}

### 解题注意事项
1、因为数组事非递减的，平方和大的数据一定在两端，通过左右指针快速找到符合要求的值
2、也可以通过冒泡排序对获取的平方进行排序

# 209_题目考察点：双指针

## 题目链接：力扣（LeetCode）官网 - 全球极客挚爱的技术成长平台

### 代码实现

package LeetCode;

public class YangSibo_209 {
    public  static void main(String args []) {
        int [] nums = {2,3,1,2,4,3};
        int target = 7;
        YangSibo_209_1 demo1 = new YangSibo_209_1();
        YangSibo_209_2 demo2 = new YangSibo_209_2();
        System.out.println("demo1:" + demo1.minSubArrayLen(target,nums));
        System.out.println("demo2:" + demo2.minSubArrayLen(target,nums));
    }
}
//暴力破解
class YangSibo_209_1 {
    public int minSubArrayLen(int target, int[] nums) {
        int length = nums.length;
        if (length == 0 || (length == 1 && nums[0] < target)) {
            return 0;
        }
        if (length == 1 && nums[0] >= target) {
            return 1;
        }
        int minres = length;
        int flag = 0;
        for (int i = 0; i < length; i++) {
            for(int j = 0 ; j < length; j++) {
                int step = j - i + 1;
                if (step < 1) {
                    continue;
                }
                int stepsum = 0;
                for (int k = i; k <= j; k++){
                    stepsum = stepsum + nums[k];
                }
                if (stepsum >= target ) {
                    flag = 1;
                    if ( step < minres ){
                        minres = step;
                    }
                }
            }
        }
        if ( flag != 0) {
            return minres;
        }else {
            return 0;
        }
    }
}

class YangSibo_209_2 {
    public int minSubArrayLen(int target, int[] nums) {
        int length = nums.length;
        if (length == 0 || (length == 1 && nums[0] < target)) {
            return 0;
        }
        if (length == 1 && nums[0] >= target) {
            return 1;
        }
        int stepsum = 0;
        int minres = length;
        int flag = 0;
        for (int i = 0, j = 0; j < length; j++) {
            stepsum = stepsum + nums[j];
            while (stepsum >= target) {
                flag = 1;
                int step = j - i + 1;
                minres = Math.min(step,minres);
                stepsum = stepsum - nums[i];
                i = i + 1;
            }
        }
        if ( flag != 0) {
            return minres;
        }else {
            return 0;
        }
    }
}

### 解题注意事项
1、暴力解法就是遍历所有存在的数组区间，记得求的数据和
2、滑动窗口的方法是通过窗口的终点这一个变量，控制数据的走向，通过逐步挪动窗口的起点找寻更优的结果

# 59_题目考察点：二分法的区间定义

## 题目链接：力扣（LeetCode）官网 - 全球极客挚爱的技术成长平台

### 代码实现

package LeetCode;

public class YangSibo_59 {
    public static void main (String args []) {
    int n = 4;
    YangSibo_59_1 demo1 = new YangSibo_59_1();
    int demo1res [][];
    demo1res = demo1.generateMatrix(n);
    for (int i = 0; i < n; i++){
        for (int j = 0; j < n; j++ ){
            System.out.println(demo1res[i][j]);
        }
    }
    }
}

class YangSibo_59_1 {
    public int[][] generateMatrix(int n) {
        int res[][] = new int[n][n];
        if (n == 1) {
            res[0][0] = 1;
            return res;
        }
        int larggestnum = n * n;
        int count = 1;
        int loop = 0;
        if (n % 2 != 0) {
            int midnum = n / 2;
            res[midnum][midnum] = larggestnum;
        }
        int i = 0, j = 0;

        while (count < larggestnum) {
            for (; j < n - loop- 1; j++) {
                res[i][j] = count;
                count++;
            }
            for (; i < n - loop- 1; i++) {
                res[i][j] = count;
                count++;
            }
            if(count > larggestnum){
                break;
            }
            for (; j > loop; j--) {
                res[i][j] = count;
                count++;
            }
            for (; i > loop; i--) {
                res[i][j] = count;
                count++;
            }
            i++;
            j++;
            loop ++;
        }
        return  res;
    }
}

### 解题注意事项
1、注意奇数和偶数结束的位置不相同，且每次都是向内循环的