221. Maximal Square

dp的值选择

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public int maximalSquare(char[][] matrix) {
	if (matrix == null || matrix.length == 0) {
		return 0;
	}
	//dp[i][j]表示以ij为右下角的含1正方型的最大边长
	int res = matrix[0][0] - 48;
	int[][] dp = new int[matrix.length][matrix[0].length];
	for (int i = 0; i < matrix.length; i++) {
		dp[i][0] = matrix[i][0] - 48;
		res = Math.max(res, dp[i][0]);
	}
	for (int i = 0; i < matrix[0].length; i++) {
		dp[0][i] = matrix[0][i] - 48;
		res = Math.max(res, dp[0][i]);
	}
	for (int i = 1; i < matrix.length; i++) {
		for (int j = 1; j < matrix[0].length; j++) {
			if (matrix[i][j] == '0') {
				dp[i][j] = 0;
			} else {
				//tmp是最该的覆盖区域
				int tmp = Math.min(dp[i - 1][j], dp[i][j - 1]);
				//matrix[i - tmp][j - tmp]-48表示该形状左上角的数字
				dp[i][j] = tmp + matrix[i - tmp][j - tmp]-48;
			}
			res = Math.max(res, dp[i][j]);
		}
	}
	return res*res;
}