[LeetCode 85]Maximal Rectangle (华为2015机试）

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题目链接：maximal-rectangle

import java.util.Arrays;/** * Given a 2D binary matrix filled with 0's and 1's,  * find the largest rectangle containing all ones and return its area. * */public class MaximalRectangle {//65 / 65 test cases passed.//Status: Accepted//Runtime: 292 ms//Submitted: 0 minutes ago//时间复杂度O(n ^ 2), 空间复杂度 O(n)    public int maximalRectangle(char[][] matrix) {    if(matrix.length == 0) {        return 0;        }    int m = matrix.length;    int n = matrix[0].length;        int[] H = new int[n];        int[] L = new int[n];        int[] R = new int[n];        Arrays.fill(H, 0);        Arrays.fill(L, 0);        Arrays.fill(R, n);                int ret = 0;        for(int i = 0; i < m; ++i) {        int left = 0;        int right = n;                for (int j = 0; j < n; j++) {if (matrix[i][j] == '1') {++H[j];L[j] = Math.max(L[j], left);} else {left = j + 1;H[j] = 0;L[j] = 0;R[j] = n;} }                for (int j = n - 1; j >= 0; j --) {if (matrix[i][j] == '1') {R[j] = Math.min(R[j], right);ret = Math.max(ret, H[j] * (R[j] - L[j]));} else {right = j;} }        }        return ret;            }}

华为2015机试题大概描述：

给定一个矩阵，里面只包含‘0’和‘1’，求出给最大的正方形的边长，该正方形里面全是‘1’

    public int maximalRectangle1(char[][] matrix) {        if(matrix.length == 0) {        return 0;        }        int m = matrix.length;    int n = matrix[0].length;        int[] H = new int[n];        int[] L = new int[n];        int[] R = new int[n];        Arrays.fill(H, 0);        Arrays.fill(L, 0);        Arrays.fill(R, n);                int maxLen = 0;        for(int i = 0; i < m; ++i) {        int left = 0;        int right = n;                for (int j = 0; j < n; j++) {if (matrix[i][j] == '1') {++H[j];L[j] = Math.max(L[j], left);} else {left = j + 1;H[j] = 0;L[j] = 0;R[j] = n;} }                for (int j = n - 1; j >= 0; j --) {if (matrix[i][j] == '1') {R[j] = Math.min(R[j], right);maxLen = Math.max(maxLen, Math.min(H[j], R[j] - L[j]));} else {right = j;} }        }        return maxLen;    }

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