标签：param 复杂度 down similar href http title i++ number
Given an integer matrix A which has the following features :
n rows and m columns.i < n, A[i][0] < A[i][1] && A[i][m - 2] > A[i][m - 1].j < m, A[0][j] < A[1][j] && A[n - 2][j] > A[n - 1][j]We define a position [i, j] is a peak if:
A[i][j] > A[i + 1][j] && A[i][j] > A[i - 1][j] &&
A[i][j] > A[i][j + 1] && A[i][j] > A[i][j - 1]
Find a peak element in this matrix. Return the index of the peak.
Guarantee that there is at least one peak, and if there are multiple peaks, return any one of them.
Example 1:
Input:
[
[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]
]
Output: [1,1]
Explanation: [2,2] is also acceptable. The element at [1,1] is 41, greater than every element adjacent to it.
Example 2:
Input:
[
[1, 5, 3],
[4,10, 9],
[2, 8, 7]
]
Output: [1,1]
Explanation: There is only one peek.
Solve it in O(n+m) time.
If you come up with an algorithm that you thought it is O(nlogm) or O(mlogn), can you prove it is actually O(n+m) or propose a similar but O(n+m) algorithm?
思路：http://courses.csail.mit.edu/6.006/spring11/lectures/lec02.pdf 或者使用二分(时间复杂度为O(nlog(n)).
// O(n + m) 解法
class Solution {
/**
* @param A: An integer matrix
* @return: The index of the peak
*/
public List<Integer> find(int x1, int x2, int y1, int y2, int[][] A) {
int mid1 = x1 + (x2 - x1) / 2;
int mid2 = y1 + (y2 - y1) / 2;
int x = mid1, y = mid2;
int max = A[mid1][mid2];
for (int i = y1; i <= y2; ++i)
if (A[mid1][i] > max) {
max = A[mid1][i];
x = mid1;
y = i;
}
for (int i = x1; i <= x2; ++i)
if (A[i][mid2] > max) {
max = A[i][mid2];
x = i;
y = mid2;
}
boolean isPeak = true;
if (A[x - 1][y] > A[x][y]) {
isPeak = false;
x -= 1;
} else if (A[x + 1][y] > A[x][y]) {
isPeak = false;
x += 1;
} else if (A[x][y - 1] > A[x][y]) {
isPeak = false;
y -= 1;
} else if (A[x][y + 1] > A[x][y]) {
isPeak = false;
y += 1;
}
if (isPeak) {
return new ArrayList<Integer>(Arrays.asList(x, y));
}
if (x >= x1 && x < mid1 && y >= y1 && y < mid2) {
return find(x1, mid1 - 1, y1, mid2 - 1, A);
}
if (x >= 1 && x < mid1 && y > mid2 && y <= y2) {
return find(x1, mid1 - 1, mid2 + 1, y2, A);
}
if (x > mid1 && x <= x2 && y >= y1 && y < mid2) {
return find(mid1 + 1, x2, y1, mid2 - 1, A);
}
if (x >= mid1 && x <= x2 && y > mid2 && y <= y2) {
return find(mid1 + 1, x2, mid2 + 1, y2, A);
}
return new ArrayList<Integer>(Arrays.asList(-1, -1));
}
public List<Integer> findPeakII(int[][] A) {
// write your code here
int n = A.length;
int m = A[0].length;
return find(1, n - 2, 1, m - 2, A);
}
}
class Solution {
/**
* @param A: An integer matrix
* @return: The index of the peak
*/
public List<Integer> findPeakII(int[][] A) {
// this is the nlog(n) method
int low = 1, high = A.length - 2;
List<Integer> ans = new ArrayList<Integer>();
while (low <= high) {
int mid = (low + high) / 2;
int col = find(mid, A);
if (A[mid][col] < A[mid - 1][col]) {
high = mid - 1;
} else if (A[mid][col] < A[mid + 1][col]) {
low = mid + 1;
} else {
ans.add(mid);
ans.add(col);
break;
}
}
return ans;
}
int find(int row, int[][] A) {
int col = 0;
for (int i = 0; i < A[row].length; i++) {
if (A[row][i] > A[row][col]) {
col = i;
}
}
return col;
}
}
标签：param 复杂度 down similar href http title i++ number
原文地址：https://www.cnblogs.com/FLAGyuri/p/12078060.html
