LintCode Find Peak Element II（查找峰值II）

最新推荐文章于 2021-09-15 01:00:00 发布

jmspan

最新推荐文章于 2021-09-15 01:00:00 发布

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分类专栏：爬梯法局部最值二分法二分法查找文章标签： lintcode

本文链接：https://blog.youkuaiyun.com/jmspan/article/details/51724388

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二分法同时被 3 个专栏收录

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这是一道LintCode上的题目，要求在一个特定的整数矩阵中找到峰值元素。矩阵的特点是相邻位置的数字不同，且在指定的行和列上存在单调性。峰值定义为一个元素，其周围四个相邻位置的元素都小于它。文中提到了两种解决方案：爬山法和二分法。目标是返回矩阵中任一峰值元素的索引。

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原题网址：http://www.lintcode.com/en/problem/find-peak-element-ii/

There is an integer matrix which has the following features:

The numbers in adjacent positions are different.
The matrix has n rows and m columns.
For all i < m, A[0][i] < A[1][i] && A[n - 2][i] > A[n - 1][i].
For all j < n, A[j][0] < A[j][1] && A[j][m - 2] > A[j][m - 1].

We define a position P is a peek if:

A[j][i] > A[j+1][i] && A[j][i] > A[j-1][i] && A[j][i] > A[j][i+1] && A[j][i] > A[j][i-1]

Find a peak element in this matrix. Return the index of the peak.

Notice

The matrix may contains multiple peeks, find any of them.

Example

Given a matrix:

[
  [1 ,2 ,3 ,6 ,5],
  [16,41,23,22,6],
  [15,17,24,21,7],
  [14,18,19,20,10],
  [13,14,11,10,9]
]

return index of 41 (which is [1,1]) or index of 24 (which is [2,2])

方法一：爬梯法/爬山法/夹逼法。

class Solution {
    private int[] di = {-1, 1, 0, 0};
    private int[] dj = {0, 0, -1, 1};
    /**
     * @param A: An integer matrix
     * @return: The index of the peak
     */
    public List<Integer> findPeakII(int[][] A) {
        // write your code here
        int i = 0, j = 0;
        while (true) {
            int max = A[i][j];
            int mi = i;
            int mj = j;
            for(int k = 0; k < 4; k++) {
                int ni = i + di[k];
                int nj = j + dj[k];
                if (ni < 0 || ni >= A.length || nj < 0 || nj >= A[ni].length) continue;
                if (A[ni][nj] > max) {
                    mi = ni;
                    mj = nj;
                    max = A[ni][nj];
                }
            }
            if (mi == i && mj == j) {
                List<Integer> result = new ArrayList<>(2);
                result.add(mi);
                result.add(mj);
                return result;
            }
            i = mi;
            j = mj;
        }
    }
}

方法二：二分法。

class Solution {
    /**
     * @param A: An integer matrix
     * @return: The index of the peak
     */
    public List<Integer> findPeakII(int[][] A) {
        // write your code here
        int top = 1, bottom = A.length - 2;
        while (top <= bottom) {
            int m = (top + bottom) / 2;
            int col = maxCol(A, m);
            if (A[m][col] < A[m - 1][col]) {
                bottom = m - 1;
            } else if (A[m][col] < A[m + 1][col]) {
                top = m + 1;
            } else {
                Integer[] result = new Integer[2];
                result[0] = m;
                result[1] = col;
                return Arrays.asList(result);
            }
        }
        return null;
    }
    
    private int maxCol(int[][] A, int row) {
        int col = 1;
        for(int i = 2; i < A[row].length - 1; i++) {
            if (A[row][i] > A[row][col]) col = i;
        }
        return col;
    }
}

方法三：两个维度都采用二分法（但我还没有证明）

class Solution {
    /**
     * @param A: An integer matrix
     * @return: The index of the peak
     */
    public List<Integer> findPeakII(int[][] A) {
        // write your code here
        int top = 1, bottom = A.length - 2;
        while (top <= bottom) {
            int m = (top + bottom) / 2;
            int col = maxCol(A, m);
            if (A[m][col] < A[m - 1][col]) {
                bottom = m - 1;
            } else if (A[m][col] < A[m + 1][col]) {
                top = m + 1;
            } else {
                Integer[] result = new Integer[2];
                result[0] = m;
                result[1] = col;
                return Arrays.asList(result);
            }
        }
        return null;
    }
    
    private int maxCol(int[][] A, int row) {
        int i = 1, j = A[row].length - 2;
        while (i <= j) {
            int m = (i + j) / 2;
            if (A[row][m] < A[row][m - 1]) j = m - 1;
            else if (A[row][m] < A[row][m + 1]) i = m + 1;
            else return m;
        }
        return 0;
    }
}